chapter 6 trigonometric functions
TRANSCRIPT
550
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Chapter 6
Trigonometric Functions Section 6.1
1. 2C rπ= ; 2A rπ=
2. d r t= ⋅
3. standard position
4. central angle
5. radian
6. rθ ; 21
2r θ
7. π
8. s
t;
t
θ
9. True
10. False; rυ ω=
11.
12.
13.
14. _
15.
16.
17.
18.
19.
20.
21.
22.
23. º1 1 1
40º10 '25" 40 10 2560 60 60
(40 0.1667 0.00694)º
40.17º
= + ⋅ + ⋅ ⋅
≈ + +≈
24. º1 1 1
61º 42 '21" 61 42 2160 60 60
(61 0.7000 0.00583)º
61.71º
= + ⋅ + ⋅ ⋅
≈ + +≈
Section 6.1: Angles and Their Measure
551
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25. º1 1 1
1º 2 '3" 1 2 360 60 60
(1 0.0333 0.00083)º
1.03º
= + ⋅ + ⋅ ⋅
≈ + +≈
26. º1 1 1
73º 40 '40" 73 40 4060 60 60
(73 0.6667 0.0111)º
73.68º
= + ⋅ + ⋅ ⋅
≈ + +≈
27. º1 1 1
9º9 '9" 9 9 960 60 60
(9 0.15 0.0025)º
9.15º
= + ⋅ + ⋅ ⋅
= + +≈
28. º1 1 1
98º 22 '45" 98 22 4560 60 60
(98 0.3667 0.0125)º
98.38º
= + ⋅ + ⋅ ⋅
≈ + +≈
29. 40.32º 40º 0.32º
40º 0.32(60 ')
40º 19.2 '
40º 19 ' 0.2 '
40º 19 ' 0.2(60")
40º 19 ' 12"
40º19 '12"
= += += += + += + += + +=
30. 61.24º 61º 0.24º
61º 0.24(60 ')
61º 14.4 '
61º 14 ' 0.4 '
61º 14 ' 0.4(60")
61º 14 ' 24"
61º14 '24"
= += += += + += + += + +=
31. 18.255º 18º 0.255º
18º 0.255(60 ')
18º 15.3'
18º 15' 0.3'
18º 15' 0.3(60")
18º 15' 18"
18º15 '18"
= += += += + += + += + +=
32. 29.411º 29º 0.411º
29º 0.411(60 ')
29º 24.66 '
29º 24 ' 0.66 '
29º 0.66(60")
29º 24 ' 39.6"
29º 24 '40"
= += += += + += += + +≈
33. 19.99º 19º 0.99º
19º 0.99(60 ')
19º 59.4 '
19º 59 ' 0.4 '
19º 59 ' 0.4(60")
19º 59 ' 24"
19º 59 '24"
= += += += + += + += + +=
34. 44.01º 44º 0.01º
44º 0.01(60 ')
44º 0.6 '
44º 0 ' 0.6 '
44º 0 ' 0.6(60")
44º 0 ' 36"
44º 0 '36"
= += += += + += + += + +=
35. 30 30 radian radian180 6
π π° = ⋅ =
36. 2
120 120 radian radians180 3
π π° = ⋅ =
37. 4
240 240 radian radians180 3
π π° = ⋅ =
38. 11
330 330 radian radians180 6
π π° = ⋅ =
39. 60 60 radian radian180 3
π π− ° = − ⋅ = −
40. 30 30 radian radian180 6
π π− ° = − ⋅ = −
41. 180 180 radian radians180
π° = ⋅ = π
42. 3
270 270 radian radians180 2
π π° = ⋅ =
Chapter 6: Trigonometric Functions
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43. 3
135 135 radian radians180 4
π π− ° = − ⋅ = −
44. 5
225 225 radian radians180 4
π π− ° = − ⋅ = −
45. 90 90 radian radians180 2
π π− ° = − ⋅ = −
46. 180 180 radian radians180
π− ° = − ⋅ = −π
47. 180
degrees 603 3
π π= ⋅ = °π
48. 5 5 180
degrees 1506 6
π π= ⋅ = °π
49. 5 5 180
degrees 2254 4
π π− = − ⋅ = − °π
50. 2 2 180
degrees 1203 3
π π− = − ⋅ = − °π
51. 180
degrees 902 2
π π= ⋅ = °π
52. 180
4 4 degrees 720π = π ⋅ = °π
53. 180
degrees 1512 12
π π= ⋅ = °π
54. 5 5 180
degrees 7512 12
π π= ⋅ = °π
55. 180
degrees 902 2
π π− = − ⋅ = − °π
56. 180
degrees 180−π = −π⋅ = − °π
57. 180
degrees 306 6
π π− = − ⋅ = − °π
58. 3 3 180
degrees 1354 4
π π− = − ⋅ = − °π
59. 17
17 17 radian radian 0.30 radian180 180
π π° = ⋅ = ≈
60. 73 73 radian180
73radians
1801.27 radians
π° = ⋅
π=
≈
61. 40 40 radian180
2 radian
90.70 radian
π− ° = − ⋅
π= −
≈ −
62. 51 51 radian180
17 radian
600.89 radian
π− ° = − ⋅
π= −
≈ −
63. 125 125 radian180
25radians
362.18 radians
π° = ⋅
π=
≈
64. 350 350 radian180
35 radians
186.11 radians
π° = ⋅
π=
≈
65. 180
3.14 radians 3.14 degrees 179.91º= ⋅ ≈π
66. 180
0.75 radian 0.75 degrees 42.97º= ⋅ ≈π
67. 180
2 radians 2 degrees 114.59º= ⋅ ≈π
68. 180
3 radians 3 degrees 171.89º= ⋅ ≈π
69. 180
6.32 radians 6.32 degrees 362.11º= ⋅ ≈π
70. 180
2 radians 2 degrees 81.03º= ⋅ ≈π
Section 6.1: Angles and Their Measure
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71. 1
10 meters; radian;2
r θ= =
110 5 meters
2s rθ= = ⋅ =
72. 6 feet; 2 radian; 6 2 12 feetr s rθ θ= = = = ⋅ =
73. 1
radian; 2 feet;3
sθ = =
( )2
6 feet1/ 3
s r
sr
θ
θ
=
= = =
74. 1
radian; 6 cm;4
sθ = =
( )6
24 cm1/ 4
s r
sr
θ
θ
=
= = =
75. 5 miles; 3 miles;r s= =
30.6 radian
5
s r
s
r
θ
θ
=
= = =
76. 6 meters; 8 meters;r s= =
8 41.333 radians
6 3
s r
s
r
θ
θ
=
= = = ≈
77. 2 inches; 30º 30 radian;180 6
r θ π π= = = ⋅ =
2 1.047 inches6 3
s rθ π π= = ⋅ = ≈
78. 2
3 meters; 120º 120 radians180 3
r θ π π= = = ⋅ =
23 2 6.283 meters
3s rθ π= = ⋅ = π ≈
79. 1
10 meters; radian2
r θ= =
( )22 21 1 1 10010 =25 m
2 2 2 4A r θ = = =
80. 6 feet; 2 radiansr θ= =
( ) ( )22 21 16 2 =36 ft
2 2A r θ= =
81. 21radian; 2 ft
3Aθ = =
2
2
2
2
1
21 1
22 3
12
6
12
12 2 3 3.464 feet
A r
r
r
r
r
θ=
=
=
=
= = ≈
82. 21radian; 6 cm
4Aθ = =
2
2
2
2
1
21 1
62 4
16
8
48
48 4 3 6.928 cm
A r
r
r
r
r
θ=
=
=
=
= = ≈
83. 25 miles; 3 mir A= =
( )
2
2
1
21
3 5225
326
0.24 radian25
A r θ
θ
θ
θ
=
=
=
= =
84. 26 meters; 8 mr A= =
( )
2
2
1
21
8 62
8 18
8 40.444 radian
18 9
A r θ
θ
θ
θ
=
=
=
= = ≈
85. 2 inches; 30º 30 radian180 6
r θ π π= = = ⋅ =
( )22 21 12 1.047 in
2 2 6 3A r
π πθ = = = ≈
Chapter 6: Trigonometric Functions
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86. 2
3 meters; 120º 120 radians180 3
r θ π π= = = ⋅ =
( )22 21 1 23 =3 9.425 m
2 2 3A r
πθ π = = ≈
87. 2 feet; radians3
rπθ= =
22 2.094 feet
3 3s rθ π π= = ⋅ = ≈
( )22 21 1 22 = 2.094 ft
2 2 3 3A r
π πθ = = = ≈
88. 4 meters; radian6
rπθ= =
24 2.094 meters
6 3s rθ π π= = ⋅ = ≈
( )22 21 1 44 4.189 m
2 2 6 3A r
π πθ = = = ≈
89. 7
12 yards; 70º 70 radians180 18
r θ π π= = = ⋅ =
712 14.661 yards
18s rθ π= = ⋅ ≈
( )22 21 1 712 28 87.965 yd
2 2 18A r
πθ π = = = ≈
90. 5
9 cm; 50º 50 radian180 18
r θ π π= = = ⋅ =
59 7.854 cm
18s rθ π= = ⋅ ≈
( )22 21 1 5 459 = 35.343 cm
2 2 18 4A r
π πθ = = ≈
91. 6 inchesr = In 15 minutes,
15 1 rev 360º 90º radians
60 4 2θ π= = ⋅ = =
6 3 9.42 inches2
s rθ π= = ⋅ = π ≈
In 25 minutes, 25 5 5
rev 360º 150º radians60 12 6
θ π= = ⋅ = =
56 5 15.71 inches
6s rθ π= = ⋅ = π ≈
92. 40 inches; 20º radian9
r θ π= = =
4040 13.96 inches
9 9s rθ π π= = ⋅ = ≈
93. 4 m; 45º 45 radian180 4
r θ π π= = = ⋅ =
( )22 21 14 2 6.28 m
2 2 4A r
πθ π = = = ≈
94. 3 cm; 60º 60 radians180 3
r θ π π= = = ⋅ =
( )22 21 1 33 4.71 cm
2 2 3 2A r
π πθ = = = ≈
95. 3
30 feet; 135º 135 radians180 4
r θ π π= = = ⋅ =
( )22 21 1 3 67530 1060.29 ft
2 2 4 2A r
π πθ = = = ≈
96. 215 yards; 100 ydr A= =
( )
2
2
1
21
100 152
100 112.5
100 80.89 radian
112.5 9
A r θ
θ
θ
θ
=
=
=
= = ≈
or 8 180 160
50.939 π π
° ⋅ = ≈ °
97. 1
5 cm; 20 seconds; radian3
r t θ= = =
( )
( )
1/ 3 1 1 1 radian/sec
20 3 20 605 1/ 3 5 1 1
cm/sec20 3 20 12
t
s rv
t t
θω
θ
= = = ⋅ =
⋅= = = = ⋅ =
98. 2 meters; 20 seconds; 5 metersr t s= = =
( ) ( )/ 5 / 2 5 1 1 radian/sec
20 2 20 85 1
m/sec20 4
s r
t ts
vt
θω = = = = ⋅ =
= = =
Section 6.1: Angles and Their Measure
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99. 26 inches; 13 inches; 35 mi/hrd r v= = =
35 mi 5280 ft 12 in. 1 hr
hr mi ft 60 min36,960 in./min
v = ⋅ ⋅ ⋅
=
36,960 in./min
13 in.2843.08 radians/min
2843.08 rad 1 rev
min 2 rad452.5 rev/min
v
rω = =
≈
≈ ⋅π
≈
100. 15 inches; 3 rev/sec 6 rad/secr ω= = = π
15 6 in./sec 90 282.7 in/sec
in. 1ft 1mi 3600sec90 16.1 mi/hr
sec 12in. 5280ft 1hr
v r
v
ω= = ⋅ π = π ≈
= π ⋅ ⋅ ⋅ ≈
101. 3960 milesr = 35º9 ' 29º57 '
5º12 '
5.2º
5.2180
0.09076 radian
θ = −==
π= ⋅
≈
3960(0.09076) 359 miless rθ= = ≈
102. 3960 milesr = 38º 21' 30º 20 '
8º1'
8.017º
8.017180
0.1399 radian
θ = −=≈
π= ⋅
≈
3960(0.1399) 554 miless rθ= = ≈
103. 3429.5 milesr =
1 rev/day 2 radians/day radians/hr12
ω π= = π =
3429.5 898 miles/hr12
v rω π= = ⋅ ≈
104. 3033.5 milesr =
1 rev/day 2 radians/day radians/hr12
ω π= = π =
3033.5 794 miles/hr12
v rω π= = ⋅ ≈
105. 52.39 10 milesr = ×
( )5
1 rev/27.3 days
2 radians/27.3 days
radians/hr12 27.3
2.39 10 2292 miles/hr327.6
v r
ω
ω
== π
π=⋅
π= = × ⋅ ≈
106. 79.29 10 milesr = ×
( )7
1 rev/365 days
2 radians/365 days
radians/hr12 365
9.29 10 66,633 miles/hr4380
v r
ω
ω
== π
π=⋅
π= = × ⋅ ≈
107. 1 22 inches; 8 inches;r r= =
1 3 rev/min 6 radians/minω π= =
Find 2ω :
1 2
1 1 2 2
2
2
2(6 ) 8
12
81.5 radians/min
1.5 rev/min
23
rev/min4
v v
r rω ωω
ω
==
π =π=
= ππ=
π
=
108. 30 feetr = 1 rev 2
0.09 radian/sec70 sec 70 sec 35
ω π π= = = ≈
rad 6 ft30 feet 2.69 feet/sec
35 sec 7 secv rω π π= = ⋅ = ≈
109. 4 feet; 10 rev/min 20 radians/minr ω= = = π
4 20
ft80
min80 ft 1 mi 60 min
min 5280 ft hr2.86 mi/hr
v rω== ⋅ π
= π
π= ⋅ ⋅
≈
Chapter 6: Trigonometric Functions
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110. 26 inches; 13 inches;d r= =
480 rev/min 960 radians/minω = = π
13 960
in12480
min12480 in 1 ft 1 mi 60 min
min 12 in 5280 ft hr37.13 mi/hr
v rω== ⋅ π
= π
π= ⋅ ⋅ ⋅
≈
80 mi/hr 12 in 5280 ft 1 hr 1 rev
13 in 1 ft 1 mi 60 min 2 rad1034.26 rev/min
v
rω =
= ⋅ ⋅ ⋅ ⋅π
≈
111. 8.5 feet; 4.25 feet; 9.55 mi/hrd r v= = =
9.55 mi/hr
4.25 ft9.55 mi 1 5280 ft 1 hr 1 rev
hr 4.25 ft mi 60 min 231.47 rev/min
v
rω = =
= ⋅ ⋅ ⋅ ⋅π
≈
112. Let t represent the time for the earth to rotate 90 miles.
24
90 2 (3559)
90(24)0.0966 hours 5.8 minutes
2 (3559)
t
t
=π
= ≈ ≈π
113. The earth makes one full rotation in 24 hours. The distance traveled in 24 hours is the circumference of the earth. At the equator the circumference is 2 (3960)π miles. Therefore,
the linear velocity a person must travel to keep up with the sun is:
2 (3960)1037 miles/hr
24
sv
t
π= = ≈
114. Find , when 3960 miles and 1'.s r θ= =
1 degree radians1' 0.00029 radian
60 min 180 degrees
3960(0.00029) 1.15 miless r
θ
θ
π= ⋅ ⋅ ≈
= = ≈
Thus, 1 nautical mile is approximately 1.15 statute miles.
115. We know that the distance between Alexandria and Syene to be 500s = miles. Since the measure of the Sun’s rays in Alexandria is 7.2° , the central angle formed at the center of Earth between Alexandria and Syene must also be 7.2° . Converting to radians, we have
7.2 7.2 radian180 25
π π° = °⋅ =°
. Therefore,
50025
25 12,500500 3979 miles
s r
r
r
θπ
π π
=
= ⋅
= ⋅ = ≈
12,5002 2 25,000C rπ π
π= = ⋅ = miles.
The radius of Earth is approximately 3979 miles, and the circumference is approximately 25,000 miles.
116. a. The length of the outfield fence is the arc length subtended by a central angle 96θ = ° with 200r = feet.
200 96 335.10 feet180
s rπθ= ⋅ = ⋅ ° ⋅ ≈
°
The outfield fence is approximately 335.1 feet long.
b. The area of the warning track is the difference between the areas of two sectors with central angle 96θ = ° . One sector with
200r = feet and the other with 190r = feet.
( )
( )
( )
2 2 2 2
2 2
1 1
2 2 296
200 1902 180
43900 3267.26
15
A R r R rθθ θ
π
π
= − = −
°= ⋅ −°
= ≈
The area of the warning track is about 3267.26 square feet.
Section 6.2: Trigonometric Functions: Unit Circle Approach
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117. 1 1 rotates at rev/minr ω , so 1 1 1v rω= .
2 2 rotates at rev/minr ω , so 2 2 2v r ω= .
Since the linear speed of the belt connecting the pulleys is the same, we have:
1 2
1 1 2 2
1 1 2 2
2 1 2 1
1 2
2 1
v v
r r
r r
r r
r
r
ω ωω ωω ω
ωω
==
=
=
118. Answers will vary.
119. If the radius of a circle is r and the length of the arc subtended by the central angle is also r, then the measure of the angle is 1 radian. Also,
1801 radian degrees
π= .
1
1360
° = revolution
120. Note that radians
1 1 0.017 radian180
π ° = ° ⋅ ≈ °
and 180
1 radian 57.296 radians
° ⋅ ≈ ° π .
Therefore, an angle whose measure is 1 radian is larger than an angle whose measure is 1 degree.
121. Linear speed measures the distance traveled per unit time, and angular speed measures the change in a central angle per unit time. In other words, linear speed describes distance traveled by a point located on the edge of a circle, and angular speed describes the turning rate of the circle itself.
122. This is a true statement. That is, since an angle measured in degrees can be converted to radian measure by using the formula 180 degrees radiansπ= , the arc length formula
can be rewritten as follows: 180
s r rπθ θ= = .
123 – 125. Answers will vary.
Section 6.2
1. 2 2 2c a b= +
2. ( ) ( )5 3 5 7 15 7 8f = − = − =
3. True
4. equal; proportional
5. 1 3
,2 2
−
6. 1
2−
7. cosine
8. ( )0,1
9. 2 2
,2 2
10. 1 3
,2 2
11. ;y x
r r
12. False
13. 3 1 3 1
, , 2 2 2 2
P x y
= = =
1sin
2t y= =
3cos
2t x= =
11 2 1 3 32
tan2 33 3 33
2
yt
x
= = = ⋅ = ⋅ =
1 1 2csc 1 2
1 12
ty
= = = ⋅ =
Chapter 6: Trigonometric Functions
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1 1 2 2 3 2 3sec 1
33 3 332
tx
= = = ⋅ = ⋅ =
32 3 2
cot 31 2 12
xt
y
= = = ⋅ =
14. 1 3 1 3
, , 2 2 2 2
P x y
= − = = −
3sin
2t y= = −
1cos
2t x= =
33 22tan = 3
1 2 12
yt
x
− = = = − ⋅ −
1 1 2 2 3 2 3csc
33 3 332
ty
= = = − = − ⋅ = −
−
1 1 2sec 1 2
1 12
tx
= = = ⋅ =
11 2 1 3 32cot2 33 3 33
2
xt
y
= = = − ⋅ = − ⋅ = −
−
15. 2 21 2 21
, , 5 5 5 5
P x y
= − = − =
21sin
5t y= =
2cos
5t x= = −
2121 5 215tan =
2 5 2 25
yt
x
= = = − −
−
1 1 5 5 21 5 21csc 1
2121 21 21215
ty
= = = ⋅ = ⋅ =
1 1 5 5sec 1
2 2 25
tx
= = = − = − −
22 55cot5 2121
5
2 21 2 21
2121 21
xt
y
− = = = − ⋅
= − ⋅ = −
16. 1 2 6 1 2 6
, , 5 5 5 5
P x y
= − = − =
2 6sin
5t y= =
1cos
5t x= = −
2 62 6 55tan = 2 6
1 5 15
yt
x
= = = − −
−
1 1 5 5 6 5 6csc 1
122 6 2 6 62 65
ty
= = = ⋅ = ⋅ =
1 1 5sec 1 5
1 15
tx
= = = − = − −
11 55cot5 2 62 6
5
1 6 6
122 6 6
xt
y
− = = = −
= − ⋅ = −
17. 2 2 2 2
, , 2 2 2 2
P x y
= − = − =
2sin
2t =
2cos
2t x= = −
Section 6.2: Trigonometric Functions: Unit Circle Approach
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22tan 1
22
yt
x
= = = −
−
1 1 2 2 2csc 1 2
2 2 222
ty
= = = ⋅ = ⋅ =
1 1 2 2 2sec 1 2
2 2 222
tx
= = = − = − ⋅ = − −
22cot 12
2
xt
y
− = = = −
18. 2 2 2 2
, , 2 2 2 2
P x y
= = =
2sin
2t y= =
2cos
2t x= =
22
tan 12
2
yt
x
= = =
1 1 2 2 2csc 1 2
2 2 222
ty
= = = ⋅ = ⋅ =
1 1 2 2 2sec 1 2
2 2 222
tx
= = = ⋅ = ⋅ =
22
cot 12
2
xt
y
= = =
19. 2 2 1 2 2 1
, , 3 3 3 3
P x y
= − = = −
1sin
3t y= = −
2 2cos
3t x= =
11 33tan3 2 22 2
3
1 2 2= =
42 2 2
yt
x
− = = = − ⋅
− ⋅ −
1 1 3csc 1 3
1 13
ty
= = = − = − −
1 1 3 3 2 3 2sec 1
42 2 2 2 22 23
tx
= = = = ⋅ =
2 22 2 33cot 2 2
1 3 13
xt
y
= = = − = −
−
20. 5 2 5 2
, , 3 3 3 3
P x y
= − − = − = −
2sin
3t y= = −
5cos
3t x= = −
22 33
tan3 55
3
2 5 2 5= =
55 5
yt
x
− = = = − − −
⋅
1 1 3 3csc 1
2 2 23
ty
= = = − = − −
Chapter 6: Trigonometric Functions
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1 1 3sec 1
553
3 5 3 5
55 5
tx
= = = − −
= − ⋅ = −
53 5 3 5
cot2 3 2 23
xt
y
− = = = − − = −
21. 11 3 8
sin sin2 2 2
3sin 4
2
3sin 2 2
2
3sin
2
1
π π π = +
π = + π
π = + ⋅ π
π =
= −
22. ( ) ( )( ) ( )
cos 7 cos 6
cos 3 2 cos 1
ππ
π = π +
= π + ⋅ = π = −
23. ( ) ( )tan 6 tan(0 6 ) tan 0 0π = + π = =
24. 7 6
cot cot2 2 2
cot 3 cot 02 2
π π π = +
π π = + π = =
25. 11 3 8
csc csc2 2 2
3csc 4
2
3csc 2 2
2
3csc
2
1
π
π
π π π = +
π = +
π = + ⋅
π =
= −
26. ( ) ( )( ) ( )
sec 8 sec 0 8
sec 0 4 2 sec 0 1
ππ
π = +
= + ⋅ = =
27. 3 3
cos cos2 2
4cos
2 2
cos ( 1) 22
cos2
0
π
π
π π − =
π = −
π = + − ⋅
π =
=
28. ( ) ( )( ) ( )
sin 3 sin 3
sin 2 sin 0π− π = − π
= − π + = − π =
29. ( ) ( )sec sec 1−π = π = −
30. ( )( ) ( )
tan 3 tan(3 )
tan 0 3 tan 0 0π− π = − π
= − + = − =
31. 2 1 1 2
sin 45º cos 60º2 2 2
++ = + =
32. 1 2 1 2
sin 30º cos 45º2 2 2
−− = − =
33. sin 90º tan 45º 1 1 2+ = + =
34. cos180º sin180º 1 0 1− = − − = −
35. 2 2 2 1
sin 45º cos 45º2 2 4 2
= ⋅ = =
36. 3 3
tan 45º cos30º 12 2
= ⋅ =
37. csc 45º tan 60º 2 3 6= ⋅ =
38. 2 3 2 3
sec30º cot 45º 13 3
= ⋅ =
39. 4sin 90º 3 tan180º 4 1 3 0 4− = ⋅ − ⋅ =
40. 5cos90º 8sin 270º 5 0 8( 1) 8− = ⋅ − − =
41. 3 3
2sin 3 tan 2 3 3 3 03 6 2 3
π π− = ⋅ − ⋅ = − =
Section 6.2: Trigonometric Functions: Unit Circle Approach
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42. 2
2sin 3 tan 2 3 1 2 34 4 2
π π+ = ⋅ + ⋅ = +
43. 3 4 3
2sec 4cot 2 2 4 2 24 3 3 3
π π+ = ⋅ + ⋅ = +
44. 2 3
3csc cot 3 1 2 3 13 4 3
π π+ = ⋅ + = +
45. csc cot 1 0 12 2
π π+ = + =
46. sec csc 1 1 22
ππ − = − − = −
47. The point on the unit circle that corresponds to
2120º
3θ π= = is
1 3,
2 2
−
.
2 3sin
3 2
π =
2 1cos
3 2
π = −
32 3 22tan 3
13 2 12
π = = ⋅ − = − −
2 1 2 3 2 3csc
3 33 332
π = = ⋅ =
2 1 2cos 1 2
13 12
π = = − = − −
12 1 2 1 3 32cot3 2 33 3 33
2
− π = = − ⋅ = − ⋅ = −
48. The point on the unit circle that corresponds to
5150º
6θ π= = is
3 1,
2 2
−
.
5 1sin
6 2
π =
5 3cos
6 2
π = −
15 1 2 3 32tan6 2 33 33
2
π = = ⋅ − ⋅ = − −
5 1 2csc 1 2
16 12
π = = ⋅ =
5 1 2 3 2 3sec 1
6 33 332
π = = ⋅ − ⋅ = − −
35 3 22cot 3
16 2 12
− π = = − ⋅ = −
49. The point on the unit circle that corresponds to
7210º
6θ π= = is
3 1,
2 2
− −
.
1sin 210º
2= −
3cos 210º
2= −
11 2 3 32
tan 210º2 33 33
2
− = = − ⋅ − ⋅ = −
1 2csc 210º 1 2
1 12
= = ⋅ − = − −
1 2 3 2 3sec 210º 1
33 332
= = ⋅ − ⋅ = − −
32 3 2
cot 210º 31 2 12
− = = − ⋅ − = −
50. The point on the unit circle that corresponds to
4240º
3θ π= = is
1 3,
2 2
− −
.
3sin 240º
2= −
1cos 240º
2= −
Chapter 6: Trigonometric Functions
562
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
33 22tan 240º 3
1 2 12
− = = − ⋅ − = −
1 2 3 2 3csc 240º 1
33 332
= = ⋅ − ⋅ = − −
1 2sec 240º 1 2
1 12
= = ⋅ − = − −
11 2 3 32cot 240º2 33 33
2
− = = − ⋅ − ⋅ = −
51. The point on the unit circle that corresponds to
3135º
4θ π= = is
2 2,
2 2
−
.
3 2sin
4 2
π =
3 2cos
4 2
π = −
23 2 22tan 14 2 22
2
π = = ⋅ − = − −
3 1 2 2 2 2csc 1 2
4 22 222
π = = ⋅ ⋅ = =
3 1 2 2 2 2sec 1 2
4 22 222
π = = ⋅ − ⋅ = − = − −
23 2 22cot 14 2 22
2
− π = = − ⋅ = −
52. The point on the unit circle that corresponds to
11495º
4θ π= = is
2 2,
2 2
−
.
11 2sin
4 2
π =
11 2cos
4 2
π = −
211 2 22tan 1
4 2 222
π = = ⋅ − = − −
11 1 2 2 2 2csc 1 2
4 22 222
π = = ⋅ ⋅ = =
11 1 2 2 2 2sec 1 2
4 22 222
π = = ⋅ − ⋅ = − = − −
211 2 22cot 1
4 2 222
− π = = − ⋅ = −
53. The point on the unit circle that corresponds to
8480º
3θ π= = is
1 3,
2 2
−
.
8 3sin
3 2
π =
8 1cos
3 2
π = −
38 3 22tan 3
13 2 12
π = = ⋅ − = − −
8 1 2 3 2 3csc 1
3 33 332
π = = ⋅ ⋅ =
8 1 2sec 1 2
13 12
π = = ⋅ − = − −
18 1 2 3 32cot3 2 33 33
2
− π = = − ⋅ ⋅ = −
Section 6.2: Trigonometric Functions: Unit Circle Approach
563
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
54. The point on the unit circle that corresponds to
13390º
6θ π= = is
3 1,
2 2
.
13 1sin
6 2
π =
13 3cos
6 2
π =
113 1 2 3 32
tan6 2 33 33
2
π = = ⋅ ⋅ =
13 1 2csc 1 2
16 12
π = = ⋅ =
13 1 2 3 2 3sec 1
6 33 332
π = = ⋅ ⋅ =
3213 3 2
cot 316 2 12
π = = ⋅ =
55. The point on the unit circle that corresponds to
9405º
4θ π= = is
2 2,
2 2
.
2sin 405º
2=
2cos 405º
2=
22 22tan 405º 1
2 222
= = ⋅ =
1 2 2csc 405º 1 2
2 222
= = ⋅ ⋅ =
1 2 2sec 405º 1 2
2 222
= = ⋅ ⋅ =
22cot 405º 12
2
= =
56. The point on the unit circle that corresponds to
13390º
6θ π= = is
3 1,
2 2
.
1sin 390º
2=
3cos390º
2=
11 2 3 32tan 390º2 33 33
2
= = ⋅ ⋅ =
1 2csc390º 1 2
1 12
= = ⋅ =
1 2 3 2 3sec390º 1
33 332
= = ⋅ ⋅ =
33 22cot 390º 3
1 2 12
= = ⋅ =
57. The point on the unit circle that corresponds to
= 30º6
θ π= − − is 3 1
,2 2
−
.
1sin
6 2
π − = −
3cos
6 2
π − =
11 2 3 32
tan6 2 33 33
2
− π − = = − ⋅ ⋅ = −
1csc = 2
162
π − − −
1 2 3 2 3sec
6 33 332
π − = = ⋅ =
33 22cot 3
16 2 12
π − = = ⋅ − = − −
Chapter 6: Trigonometric Functions
564
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58. The point on the unit circle that corresponds to
60º3
θ π= − = − is 1 3
,2 2
−
.
3sin
3 2
π − = −
1cos
3 2
π − =
33 22tan 3
13 2 12
− π − = = − ⋅ = −
1 2 3 2 3csc 1
3 33 332
π − = = ⋅ − ⋅ = − −
1 2sec 1 2
13 12
π − = = ⋅ =
11 2 3 32
cot3 2 33 33
2
π − = = ⋅ − ⋅ = − −
59. The point on the unit circle that corresponds to
3135º
4θ π= − = − is
2 2,
2 2
− −
.
( ) 2sin 135º
2− = −
( ) 2cos 135º
2− = −
( )2
2 22tan 135º 12 22
2
− − = = − ⋅ − = −
( ) 1 2 2csc 135º 1 2
2 222
− = = ⋅ − ⋅ = − −
( ) 1 2 2sec 135º 1 2
2 222
− = = ⋅ − ⋅ = −
( )2
2cot 135º 12
2
− − = =
−
60. The point on the unit circle that corresponds to
4240º
3
πθ = − = − is 1 3
,2 2
−
.
( ) 3sin 240º
2− =
( ) 1cos 240º
2− = −
( )3
3 22tan 240º 31 2 12
− = = ⋅ − = − −
( ) 1 2 3 2 3csc 240º 1
33 332
− = = ⋅ ⋅ =
( ) 1 2sec 240º 1 2
1 12
− = = ⋅ − = − −
( )1
1 2 3 32cot 240º2 33 33
2
− − = = − ⋅ ⋅ = −
61. The point on the unit circle that corresponds to 5
450º2
πθ = = is ( )0, 1 .
5sin 1
2
π = 5 1
csc 12 1
π = =
5cos 0
2
π = 5 1
sec undefined2 0
π = =
5 1tan undefined
2 0
π = = 5 0
cot 02 1
π = =
62. The point on the unit circle that corresponds to 5 900ºθ π= = is ( )1, 0− .
sin 5 0π = 1
csc5 undefined0
π = =
cos5 1π = − 1
sec5 11
π = = −−
0tan 5 0
1π = =
−
1cot 5 undefined
0π −= =
Section 6.2: Trigonometric Functions: Unit Circle Approach
565
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
63. The point on the unit circle that corresponds to
14840
3
πθ = − = − ° is 1 3
,2 2
− −
.
14 3 14 1sin cos
3 2 3 2
π π − = − − = −
314 3 22tan 3
13 2 12
π
− − = = − ⋅ − = −
14 1 2 3 2 3csc 1
3 33 332
π − = = ⋅ − ⋅ = − −
14 1 2sec 1 2
13 12
π − = = ⋅ − = − −
114 1 2 3 32cot
3 2 33 332
π − − = − ⋅ − ⋅ = −
64. The point on the unit circle that corresponds to
13390
6
πθ = − = − ° is 3 1
,2 2
−
.
13 1 13 3sin cos
6 2 6 2
π π − = − − =
113 1 2 3 32
tan6 2 33 33
2
− π − = = − ⋅ ⋅ = −
13 1csc = 2
162
π − − −
13 1 2 3 2 3sec
6 33 332
π − = = ⋅ =
313 3 22cot 3
16 2 12
π − = = ⋅ − = − −
65. Set the calculator to degree mode: sin 28º 0.47≈ .
66. Set the calculator to degree mode: cos14º 0.97≈ .
67. Set the calculator to degree mode: 1
sec 21º 1.07cos 21
= š
.
68. Set the calculator to degree mode: 1
cot 70º 0.36tan 70º
= ≈ .
69. Set the calculator to radian mode: tan 0.3210
π ≈ .
70. Set the calculator to radian mode: sin 0.388
π ≈ .
Chapter 6: Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
71. Set the calculator to radian mode: 1
cot 3.7312 tan
12
π = ≈π
.
72. Set the calculator to radian mode: 5 1
csc 1.07513 sin13
π = ≈π
.
73. Set the calculator to radian mode: sin 1 0.84≈ .
74. Set the calculator to radian mode: tan 1 1.56≈ .
75. Set the calculator to degree mode: sin1º 0.02≈ .
76. Set the calculator to degree mode: tan1º 0.02≈ .
77. For the point ( 3, 4)− , 3x = − , 4y = ,
2 2 9 16 25 5r x y= + = + = =
4 5sin csc
5 43 5
cos sec5 34 3
tan cot3 4
θ θ
θ θ
θ θ
= =
= − = −
= − = −
78. For the point (5, 12)− , 5x = , 12y = − ,
2 2 25 144 169 13r x y= + = + = =
12 13sin csc
13 125 13
cos sec13 5
12 5tan cot
5 12
θ θ
θ θ
θ θ
= − = −
= =
= − = −
79. For the point (2, 3)− , 2x = , 3y = − ,
2 2 4 9 13r x y= + = + =
3 13 3 13 13sin csc
13 313 13
2 13 2 13 13cos sec
13 213 133 2
tan cot2 3
θ θ
θ θ
θ θ
−= ⋅ = − = −
= ⋅ = =
= − = −
80. For the point ( 1, 2)− − , 1x = − , 2y = − ,
2 2 1 4 5r x y= + = + =
2 5 2 5 5 5sin csc
5 2 25 5
1 5 5 5cos sec 5
5 15 5
2 1 1tan 2 cot
1 2 2
θ θ
θ θ
θ θ
−= ⋅ = − = = −−
−= ⋅ = − = = −−
− −= = = =− −
81. For the point ( 2, 2)− − , 2x = − , 2y = − ,
2 2 4 4 8 2 2r x y= + = + = =
2 2 2 2 2sin csc 2
2 22 2 2
2 2 2 2 2cos sec 2
2 22 2 2
2 2tan 1 cot 1
2 2
θ θ
θ θ
θ θ
−= ⋅ = − = = −
−
−= ⋅ = − = = −
−
− −= = = =
− −
Section 6.2: Trigonometric Functions: Unit Circle Approach
567
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
82. For the point ( 1, 1)− , 1x = − , 1y = ,
2 2 1 1 2r x y= + = + =
1 2 2 2sin csc 2
2 12 2
1 2 2 2cos sec 2
2 12 21 1
tan 1 cot 11 1
θ θ
θ θ
θ θ
= ⋅ = = =
−= ⋅ = − = = −−−= = − = = −
−
83. For the point 1 1
,3 4
, 1
3x = ,
1
4y = ,
2 2 1 1 25 5
9 16 144 12r x y= + = + = =
11 12 34sin
5 4 5 512
θ = = ⋅ =
55 4 512csc
1 12 1 34
θ = = ⋅ =
11 12 43cos
5 3 5 512
θ = = ⋅ =
55 3 512sec
1 12 1 43
θ = = ⋅ =
11 3 34tan
1 4 1 43
θ = = ⋅ =
11 4 43cot
1 3 1 34
θ = = ⋅ =
84. For the point (0.3, 0.4) , 0.3x = , 0.4y = ,
2 2 0.09 0.16 0.25 0.5r x y= + = + = =
0.4 4 0.5 5sin csc
0.5 5 0.4 40.3 3 0.5 5
cos sec0.5 5 0.3 30.4 4 0.3 3
tan cot0.3 3 0.4 4
θ θ
θ θ
θ θ
= = = =
= = = =
= = = =
85. sin 45º sin135º sin 225º sin 315º
2 2 2 2
2 2 2 2
0
+ + +
= + + − + −
=
86. 3
tan 60º tan150º 33
3 3 3 2 3
3 3
+ = + −
−= =
87.
( )( )
sin 40º sin130º sin 220º sin 310º
sin 40º sin130º sin 40º 180º
sin 130º 180º
sin 40º sin130º sin 40º sin130º
0
+ + += + + + +
+= + − −=
88. ( )tan 40º tan140º tan 40º tan 180º 40º
tan 40º tan 40º
0
+ = + −= −=
89. If ( ) sin 0.1f θ θ= = , then
( ) sin( ) 0.1f θ π θ+ = + π = − .
90. If ( ) cos 0.3f θ θ= = , then
( ) cos( ) 0.3f θ π θ+ = + π = − .
91. If ( ) tan 3f θ θ= = , then
( ) tan( ) 3f θ π θ+ = + π = .
92. If ( ) cot 2f θ θ= = − , then
( ) cot( ) 2f θ π θ+ = + π = − .
93. If 1
sin5
θ = , then 1 5
csc 1 51 15
θ = = ⋅ =
.
94. If 2
cos3
θ = , then 1 3 3
sec 12 2 23
θ = = ⋅ =
.
95. ( ) ( ) 360º sin 60º
2f = =
96. ( ) ( ) 160º cos 60º
2g = =
97. ( )60º 60º 1sin sin 30º
2 2 2f = = =
98. ( )60º 60º 3cos cos 30º
2 2 2g = = =
99. ( ) ( )2
2 2 3 360º sin 60º
2 4f
= = =
Chapter 6: Trigonometric Functions
568
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
100. ( ) ( )2
2 2 1 160º cos 60º
2 4g
= = =
101. ( ) ( ) ( ) 32 60º sin 2 60º sin 120º
2f ⋅ = ⋅ = =
102. ( ) ( ) ( ) 12 60º cos 2 60º cos 120º
2g ⋅ = ⋅ = = −
103. ( ) ( ) 32 60º 2sin 60º 2 3
2f = = ⋅ =
104. ( ) ( ) 12 60º 2cos 60º 2 1
2g = = ⋅ =
105. ( ) ( ) ( ) 360º sin 60º sin 300º
2f − = − = = −
106. ( ) ( ) ( ) 160º cos 60º cos 300º
2g − = − = =
107. ( )(30 ) (30 ) (30 )
sin 30 cos30
1 3 1 3
2 2 2
f g f g+ ° = ° + °= ° + °
+= + =
108. ( )(60 ) (60 ) (60 )
sin 60 cos 60
3 1 3 1
2 2 2
f g f g− ° = ° − °= ° − °
−= − =
109. 3 3 3
( )4 4 4
3 3sin cos
4 4
2 2
2 2
4 2 1
4 4 2
f g f gπ π π
π π
⋅ = ⋅
=
= ⋅ −
= − = − = −
110. 4 4 4
( )3 3 3
4 4sin cos
3 3
3 1 3
2 2 4
f g f gπ π π
π π
⋅ = ⋅
=
= − ⋅ − =
111. ( )6 6
sin 26
3sin
3 2
f h f hπ π
π
π
= =
= =
112. ( )( )(60 ) (60 )
60cos
2
3cos30
2
g p g p° = °
° =
= ° =
113. ( )( )(315 ) (315 )
cos315
21
cos3152
1 2 2
2 2 4
p g p g° = °°=
= °
= ⋅ =
114. 5 5
( )6 6
5 12 sin 2 1
6 2
h f h fπ π
π
= = = ⋅ =
115. a. 2
sin4 4 2
fπ π = =
The point 2
,4 2
π
is on the graph of f.
b. The point 2
,2 4
π
is on the graph of 1f − .
c. 3 34 4 2
sin 32
1 3
2
f fπ π π
π
+ − = −
= −
= −= −
The point , 24
π −
is on the graph of
34
y f xπ = + −
.
Section 6.2: Trigonometric Functions: Unit Circle Approach
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116. a. 3
cos6 6 2
gπ π = =
The point 3
,6 2
π
is on the graph of g.
b. The point 3
,2 6
π
is on the graph of 1g − .
c. 2 2 (0)6 6
2cos(0)
2 1
2
g gπ π − =
== ⋅=
Thus, the point , 26
π
is on the graph of
26
y g xπ = −
.
117. Answers will vary. One set of possible answers
is 11 5 7 13
, , , ,3 3 3 3 3
π π π π π− − .
118. Answers will vary. One ser of possible answers
is 13 5 3 11 19
, , , ,4 4 4 4 4
π π π π π− −
119.
sinsin
0.5 0.4794 0.9589
0.4 0.3894 0.9735
0.2 0.1987 0.9933
0.1 0.0998 0.9983
0.01 0.0100 1.0000
0.001 0.0010 1.0000
0.0001 0.0001 1.0000
0.00001 0.00001 1.0000
θθ θθ
( ) sinf
θθθ
= approaches 1 as θ approaches 0.
120.
cos 1cos 1
0.5 0.1224 0.2448
0.4 0.0789 0.1973
0.2 0.0199 0.0997
0.1 0.0050 0.0050
0.01 0.00005 0.0050
0.001 0.0000 0.0005
0.0001 0.0000 0.00005
0.00001 0.0000 0.000005
θθ θθ
−−
− −− −− −− −− −
−−
−
( ) cos 1g
θθθ
−= approaches 0 as θ
approaches 0.
121. Use the formula ( ) ( )20 sin 2v
Rg
θθ = with
232.2ft/secg = ; 45ºθ = ; 0 100 ft/secv = :
( )2(100) sin(2 45º )
45º 310.56 feet32.2
R⋅= ≈
Use the formula ( ) ( )220 sin
2
vH
g
θθ = with
232.2ft/secg = ; 45ºθ = ; 0 100 ft/secv = :
( )2 2100 (sin 45º )
45º 77.64 feet2(32.2)
H = ≈
122. Use the formula ( ) ( )20 sin 2v
Rg
θθ = with
29.8 m/secg = ; 30ºθ = ; 0 150 m/secv = :
( )2150 sin(2 30º )
30º 1988.32 m9.8
R⋅= ≈
Use the formula ( ) ( )220 sin
2
vH
g
θθ = with
29.8 m/secg = ; 30ºθ = ; 0 150 m/secv = :
( )2 2150 (sin 30º )
30º 286.99 m2(9.8)
H = ≈
123. Use the formula ( ) ( )20 sin 2v
Rg
θθ = with
29.8 m/secg = ; 25ºθ = ; 0 500 m/secv = :
( )2
19,541.95500 sin(2 25º )
25º m9.8
R⋅= ≈
Chapter 6: Trigonometric Functions
570
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Use the formula ( ) ( )220 sin
2
vH
g
θθ = with
29.8 m/secg = ; 25ºθ = ; 0 500 m/secv = :
( )2 2500 (sin 25º )
25º 2278.14 m2(9.8)
H = ≈
124. Use the formula ( ) ( )20 sin 2v
Rg
θθ = with
232.2ft/secg = ; 50ºθ = ; 0 200 ft/secv = :
( )2200 sin(2 50º )
50º 1223.36 ft32.2
R⋅= ≈
Use the formula ( ) ( )220 sin
2
vH
g
θθ = with
232.2ft/secg = ; 50ºθ = ; 0 200 ft/secv = :
( )2 2200 (sin 50º )
50º 364.49 ft2(32.2)
H = ≈
125. Use the formula ( ) 2
sin cos
at
gθ
θ θ= with
232 ft/secg = and 10 feeta = :
a. ( ) ( )2 1030 1.20 seconds
32sin 30º cos30ºt = ≈
⋅
b. ( ) ( )2 1045 1.12 seconds
32sin 45º cos 45ºt = ≈
⋅
c. ( ) ( )2 1060 1.20 seconds
32sin 60º cos 60ºt = ≈
⋅
126. Use the formula
( ) cos 16 0.5cos(2 )x θ θ θ= + + .
( ) ( )( ) ( )
30 cos 30º 16 0.5cos(2 30º )
cos 30º 16 0.5cos 60º
4.90 cm
x = + + ⋅
= + +
≈
( ) ( )( ) ( )
45 cos 45º 16 0.5cos(2 45º )
cos 45º 16 0.5cos 90º
4.71 cm
x = + + ⋅
= + +
≈
127. Note: distance on road
time on roadrate on road
=
8 2
8
14
1tan1
41
14 tan
x
x
θ
θ
−=
= −
= −
= −
a. 2 1
(30º ) 13sin 30º 4 tan 30º
2 11
1 13 4
2 3
4 31 1.9 hr
3 4
T = + −
= + −⋅ ⋅
= + − ≈
Sally is on the paved road for 1
1 0.57 hr4 tan 30º
− ≈ .
b. 2 1
(45º ) 13sin 45º 4 tan 45º
2 11
1 4 132
2 2 11 1.69 hr
3 4
T = + −
= + −⋅⋅
= + − ≈
Sally is on the paved road for
o
11 0.75 hr
4 tan 45− = .
c. o o
2 1(60º ) 1
3sin 60 4 tan 602 1
13 4 3
32
4 11
3 3 4 31.63 hr
T = + −
= + −⋅⋅
= + −
≈
Sally is on the paved road for
o
11 0.86 hr
4 tan 60− ≈ .
Section 6.2: Trigonometric Functions: Unit Circle Approach
571
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
d. 2 1
(90º ) 13sin 90º 4 tan 90º
T = + − .
But tan 90º is undefined, so we cannot use the function formula for this path. However, the distance would be 2 miles in the sand and 8 miles on the road. The total
time would be: 2 5
1 1.673 3
+ = ≈ hours. The
path would be to leave the first house walking 1 mile in the sand straight to the road. Then turn and walk 8 miles on the road. Finally, turn and walk 1 mile in the sand to the second house.
128. When 30ºθ = :
( ) ( )( )
3
3 32
1 sec30º130º (2) 251.42 cm
3 tan 30ºV π
+= ≈
When 45ºθ = :
( ) ( )( )
3
3 32
1 sec 45º145º (2) 117.88 cm
3 tan 45ºV π
+= ≈
When 60ºθ = :
( ) ( )( )
3
3 32
1 sec60º160º (2) 75.40 cm
3 tan 60ºV π
+= ≈
129. a. ( ) ( ) ( )[ ]
( ) ( )[ ]
2
2
32 260 sin 2 60º cos 2 60º 1
32
32 2sin 120º cos 120º 1
32
32 2 1
16.56 ft
3 1
2 2
R = ⋅ − ⋅ −
= − −
≈ − −
≈
−
b. Let ( ) ( )2
1
32 2sin 2 cos 2 1
32Y x x= − −
0
20
45° 90°
c. Using the MAXIMUM feature, we find:
0
20
90°45°
is largest when 67.5ºR θ = .
130. sin 0 sin
Slope of tancos 0 cos
Mθ θ θθ θ
−= = =−
.
Since L is parallel to M, the slope of L is equal to the slope of M. Thus, the slope of tanL θ= .
131. a. When 1t = , the coordinate on the unit circle is approximately (0.5, 0.8) . Thus,
sin1 0.8≈ 1
csc1 1.30.8
≈ ≈
cos1 0.5≈ 1
sec1 2.00.5
≈ =
0.8tan1 1.6
0.5≈ =
0.5cot1 0.6
0.8≈ ≈
Set the calculator on RADIAN mode:
b. When 5.1t = , the coordinate on the unit circle is approximately (0.4, 0.9)− . Thus,
sin 5.1 0.9≈ − 1
csc5.1 1.10.9
≈ ≈ −−
cos5.1 0.4≈ 1
sec5.1 2.50.4
≈ =
0.9tan 5.1 2.3
0.4
−≈ ≈ − 0.4
cot 5.1 0.40.9
≈ ≈ −−
Set the calculator on RADIAN mode:
132. a. When 2t = , the coordinate on the unit circle is approximately ( 0.4, 0.9)− . Thus,
sin 2 0.9≈ 1
csc 2 1.10.9
≈ ≈
Chapter 6: Trigonometric Functions
572
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
cos 2 0.4≈ − 1
sec 2 2.50.4
≈ = −−
0.9tan 2 2.3
0.4≈ = −
−
0.4cot 2 0.4
0.9
−≈ ≈ −
Set the calculator on RADIAN mode:
b. When 4t = , the coordinate on the unit circle is approximately ( 0.7, 0.8)− − . Thus,
sin 4 0.8≈ − 1
csc 4 1.30.8
≈ ≈ −−
cos 4 0.7≈ − 1
sec 4 1.40.7
≈ ≈ −−
0.8tan 4 1.1
0.7
−≈ ≈−
0.7
cot 4 0.90.8
−≈ ≈−
Set the calculator on RADIAN mode:
133 – 135. Answers will vary.
Section 6.3
1. All real numbers except 1
2− ;
1|
2x x
≠ −
2. even
3. False
4. True
5. 2π , π
6. All real number, except odd multiples of 2
π
7. All real numbers between 1− and 1, inclusive. That is, { }| 1 1y y− ≤ ≤ or [ ]1, 1− .
8. True
9. 1
10. False; 1
seccos
θθ
=
11. 2
sin 405º sin(360º 45º ) sin 45º2
= + = =
12. 1
cos 420º cos(360º 60º ) cos 60º2
= + = =
13. tan 405º tan(180º 180º 45º ) tan 45º 1= + + = =
14. 1
sin 390º sin(360º 30º ) sin 30º2
= + = =
15. csc 450º csc(360º 90º ) csc90º 1= + = =
16. sec540º sec(360º 180º ) sec180º 1= + = = −
17. cot 390º cot(180º 180º 30º ) cot 30º 3= + + = =
18. sec 420º sec(360º 60º ) sec60º 2= + = =
19. 33
cos cos 84 4
π π = + π
cos 4 24
cos4
2
2
π = + ⋅ π π=
=
20. 9 2
sin sin 2 sin4 4 4 2
π π π = + π = =
21. ( ) ( )tan 21 tan(0 21 ) tan 0 0π = + π = =
22. 9
csc csc 42 2
π π = + π
csc 2 22
csc2
1
π = + ⋅ π π=
=
23. 17
sec sec 44 4
π π = + π
sec 2 24
sec4
2
π = + ⋅ π π=
=
Section 6.3: Properties of the Trigonometric Functions
573
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
24. 17
cot cot 44 4
π π = + π
cot 2 24
cot4
1
π = + ⋅ π π=
=
25. 19 3
tan tan 3 tan6 6 6 3
π π π = + π = =
26. 25
sec sec 46 6
π π = + π
sec 2 26
sec6
2 3
3
π = + ⋅ π π=
=
27. Since sin 0θ > for points in quadrants I and II, and cos 0θ < for points in quadrants II and III, the angle θ lies in quadrant II.
28. Since sin 0θ < for points in quadrants III and IV, and cos 0θ > for points in quadrants I and IV, the angle θ lies in quadrant IV.
29. Since sin 0θ < for points in quadrants III and IV, and tan 0θ < for points in quadrants II and IV, the angle θ lies in quadrant IV.
30. Since cos 0θ > for points in quadrants I and IV, and tan 0θ > for points in quadrants I and III, the angle θ lies in quadrant I.
31. Since cos 0θ > for points in quadrants I and IV, and tan 0θ < for points in quadrants II and IV, the angle θ lies in quadrant IV.
32. Since cos 0θ < for points in quadrants II and III, and tan 0θ > for points in quadrants I and III, the angle θ lies in quadrant III.
33. Since sec 0θ < for points in quadrants II and III, and sin 0θ > for points in quadrants I and II, the angle θ lies in quadrant II.
34. Since csc 0θ > for points in quadrants I and II, and cos 0θ < for points in quadrants II and III, the angle θ lies in quadrant II.
35. 3 4
sin , cos5 5
θ θ= − =
3sin 3 5 35
tan4cos 5 4 45
θθθ
− = = = − ⋅ = −
1 1 5csc
3sin 35
θθ
= = = − −
1 1 5sec
4cos 45
θθ
= = =
1 4cot
tan 3θ
θ= = −
36. 4 3
sin , cos5 5
θ θ= = −
4sin 4 5 45
tan3cos 5 3 35
θθθ
= = = ⋅ − = − −
1 1 5csc
4sin 45
θθ
= = =
1 1 5sec
3cos 35
θθ
= = = − −
1 3cot
tan 4θ
θ= = −
37. 2 5 5
sin , cos5 5
θ θ= =
2 55sin 2 5 5
tan 2cos 5 55
5
θθθ
= = = ⋅ =
1 1 5 5 5csc 1
sin 22 5 52 55
θθ
= = = ⋅ ⋅ =
Chapter 6: Trigonometric Functions
574
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
1 1 5 5sec 5
cos 5 555
θθ
= = = ⋅ =
1 1cot
tan 2θ
θ= =
38. 5 2 5
sin , cos5 5
θ θ= − = −
5
5 5 5 1
5 22 52 5
5
sintan
cos
θθθ
−
= − ⋅ − =
−
= =
1 1 5 5csc 1 5
sin 5 555
θθ
= = = ⋅ − ⋅ = − −
1 1 5 5 5sec
cos 22 5 52 55
θθ
= = = − ⋅ = − −
1 1 2
cot 1 21tan 12
θθ
= = = ⋅ =
39. 1 3
sin , cos2 2
θ θ= =
1sin 1 2 3 32
tancos 2 33 33
2
θθθ
= = = ⋅ ⋅ =
1 1 2csc 1 2
1sin 12
θθ
= = = ⋅ =
1 1 2 3 2 3sec
cos 33 332
θθ
= = = ⋅ =
1 1 3 3cot 3
tan 3 333
θθ
= = = ⋅ =
40. 3 1
sin , cos2 2
θ θ= =
32sin 3 2
tan 31cos 2 12
θθθ
= = = ⋅ =
1 1 2 3 2 3csc 1
sin 33 332
θθ
= = = ⋅ ⋅ =
1 1 2sec 1 2
1cos 12
θθ
= = = ⋅ =
1 1 1 3 3cot
tan 33 3 3θ
θ= = = ⋅ =
41. 1 2 2
sin , cos3 3
θ θ= − =
1sin 1 3 2 23
tancos 3 42 2 22 2
3
θθθ
− = = = − ⋅ ⋅ = −
1 1 3csc 1 3
1sin 13
θθ
= = = ⋅ − = − −
1 1 3 2 3 2sec
cos 42 2 22 23
θθ
= = = ⋅ =
1 1 4 2cot 2 2
tan 2 224
θθ
= = = − ⋅ = −
−
42. 2 2 1
sin , cos3 3
θ θ= = −
2 23sin 2 2 3
tan 2 21cos 3 13
θθθ
= = = ⋅ − = − −
1 1 3 2 3 2csc 1
sin 42 2 22 23
θθ
= = = ⋅ ⋅ =
Section 6.3: Properties of the Trigonometric Functions
575
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
1 1 3sec 1 3
1cos 13
θθ
= = = ⋅ − = − −
1 1 1 2 2cot
tan 42 2 2 2 2θ
θ= = = − ⋅ = −
−
43. 12
sin , in quadrant II13
θ θ=
Solve for cosθ : 2 2
2 2
2
sin cos 1
cos 1 sin
cos 1 sin
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant II, cos 0θ θ < .
2
2
cos 1 sin
12 144 25 51 1
13 169 169 13
θ θ= − −
= − − = − − = − = −
12sin 12 13 1213
tan5cos 13 5 5
13
θθθ
= = = ⋅ − = − −
1 1 13csc
12sin 1213
θθ
= = =
1 1 13sec
5cos 513
θθ
= = = − −
1 1 5cot
12tan 125
θθ
= = = − −
44. 3
cos , in quadrant IV5
θ θ=
Solve for sinθ : 2 2
2 2
2
sin cos 1
sin 1 cos
sin 1 cos
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant IV, sin 0θ θ < . 2
2 3 16 4sin 1 cos 1
5 25 5θ θ = − − = − − = − = −
4sin 4 5 45
tan3cos 5 3 35
θθθ
− = = = − ⋅ = −
1 1 5csc
4sin 45
θθ
= = = − −
1 1 5sec
3cos 35
θθ
= = =
1 1 3cot
4tan 43
θθ
= = = − −
45. 4
cos , in quadrant III5
θ θ= −
Solve for sinθ : 2 2
2 2
2
sin cos 1
sin 1 cos
sin 1 cos
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant III, sin 0θ θ < .
2
2
sin 1 cos
4 16 9 31 1
5 25 25 5
θ θ= − −
= − − − = − − = − = −
3sin 3 5 35
tan4cos 5 4 45
θθθ
− = = = − ⋅ − = −
1 1 5csc
3sin 35
θθ
= = = − −
1 1 5sec
4cos 45
θθ
= = = − −
1 1 4cot
3tan 34
θθ
= = =
46. 5
sin , in quadrant III13
θ θ= −
Solve for cosθ : 2 2
2 2
2
sin cos 1
cos 1 sin
cos 1 sin
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant III, cos 0θ θ < . 2
2 5cos 1 sin 1
13
144 12
169 13
θ θ = − − = − − −
= − = −
Chapter 6: Trigonometric Functions
576
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
5sin 5 13 513tan
12cos 13 12 1213
θθθ
− = = = − ⋅ − = −
1 1 13csc
5sin 513
θθ
= = = − −
1 1 13sec
12cos 1213
θθ
= = = − −
1 1 12cot
5tan 512
θθ
= = =
47. 5
sin , 90º 180º , in quadrant II13
θ θ θ= < <
Solve for cosθ : 2 2
2 2
2
sin cos 1
cos 1 sin
cos 1 sin
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant II, cos 0θ θ < . 2
2 5cos 1 sin 1
13
25 144 121
169 169 13
θ θ = − − = − −
= − − = − = −
5sin 5 13 513
tan12cos 13 12 1213
θθθ
= = = ⋅ − = − −
1 1 13csc
5sin 513
θθ
= = =
1 1 13sec
12cos 1213
θθ
= = = − −
1 1 12cot
5tan 512
θθ
= = = − −
48. 4
cos , 270º 360º ; in quadrant IV5
θ θ θ= < <
Solve for sinθ : 2 2
2
sin cos 1
sin 1 cos
θ θ
θ θ
+ =
= ± −
Since is in quadrant IV, sin 0θ θ < .
2
2 4 9 31
5 25 5sin 1 cosθ θ = − − = − = − = − −
3sin 3 5 35
tan4cos 5 4 45
θθθ
− = = = − ⋅ = −
1 1 5csc
3sin 35
θθ
= = = − −
1 1 5sec
4cos 45
θθ
= = =
1 1 4cot
3tan 34
θθ
= = = − −
49. 1
cos , , in quadrant II3 2
πθ θ θ= − < < π
Solve for sinθ : 2 2
2 2
2
sin cos 1
sin 1 cos
sin 1 cos
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant II, sin 0θ θ > .
2
2
sin 1 cos
1 1 8 2 21 1
3 9 9 3
θ θ= −
= − − = − = =
2 23sin 2 2 3
tan 2 21cos 3 13
1 1 3 2 3 2csc
sin 42 2 22 23
θθθ
θθ
= = = ⋅ − = − −
= = = ⋅ =
1 1sec 3
1cos3
1 1 2 2cot
tan 42 2 2
θθ
θθ
= = = − −
= = ⋅ = −−
50. 2 3
sin , , in quadrant III3 2
πθ θ θ= − π < <
Solve for cosθ : 2 2
2
sin cos 1
cos 1 sin
θ θ
θ θ
+ =
= ± −
Since is in quadrant III, cos 0θ θ < .
Section 6.3: Properties of the Trigonometric Functions
577
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
22 2
cos 1 sin 13
5 5
9 3
θ θ = − − = − − −
= − = −
2sin 3
tancos 5
3
2 3 5 2 5
3 55 5
θθθ
− = =
−
= − ⋅ − ⋅ =
1csc
sinθ
θ= 1 3
2 23
= = − −
1sec
cosθ
θ= 1 3 5 3 5
55 553
= = − ⋅ = −
−
1cot
tanθ
θ= = 1 5 5 5
22 5 52 55
= ⋅ =
51. 2
sin , tan 0, so is in quadrant II3
θ θ θ= <
Solve for cosθ : 2 2
2 2
2
sin cos 1
cos 1 sin
cos 1 sin
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant II, cos 0θ θ < .
2
2
cos 1 sin
2 4 5 51 1
3 9 9 3
θ θ= − −
= − − = − − = − = −
2sin 2 3 2 53
tancos 3 555
3
1 1 3csc
2sin 23
θθθ
θθ
= = = ⋅ − = − −
= = =
1 1 3 3 5sec
cos 5553
1 1 5 5cottan 22 52 5
5
θθ
θθ
= = = − = −
−
= = = − = − −
52. 1
cos , tan 04
θ θ= − >
Since sin
tan 0cos
θθθ
= > and cos 0θ < , sin 0θ < .
Solve for sinθ : 2 2
2
sin cos 1
sin 1 cos
θ θ
θ θ
+ =
= ± −
2
2
sin 1 cos
1 11 1
4 16
15 15
16 4
θ θ= − −
= − − − = − −
= − = −
sintan
cos
θθθ
= =
154 15 41 4 14
− = − ⋅ − −
15=
1 1 4 15 4 15csc
sin 1515 15154
θθ
= = = − ⋅ = −
−
1sec
cosθ
θ= = 1
14
−
44
1= − = −
1 1 15 15cot
tan 1515 15θ
θ= = ⋅ =
Chapter 6: Trigonometric Functions
578
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
53. sec 2, sin 0, so is in quadrant IVθ θ θ= <
Solve for cosθ : 1 1
cossec 2
θθ
= =
Solve for sinθ : 2 2
2 2
2
sin cos 1
sin 1 cos
sin 1 cos
θ θθ θ
θ θ
+ == −
= ± −
Since is in quadrant IV, sin 0θ θ < .
2
2
sin 1 cos
1 1 3 31 1
2 4 4 2
θ θ= − −
= − − = − − = − = −
3sin 3 22tan 3
1cos 2 12
1 1 2 2 3csc
sin 3332
θθθ
θθ
− = = = − ⋅ = −
= = = − = −
−
1 1 3cot
tan 33θ
θ= = = −
−
54. csc 3, cot 0, so is in quadrant IIθ θ θ= <
Solve for sinθ : 1 1
sincsc 3
θθ
= =
Solve for cosθ : 2 2
2
sin cos 1
cos 1 sin
θ θ
θ θ
+ =
= ± −
Since is in quadrant II, cos 0θ θ < .
2
2
cos 1 sin
1 1 8 2 21 1
3 9 9 3
θ θ= − −
= − − = − − = − = −
1sin 3tancos 2 2
3
1 3 2 2
3 42 2 2
θθθ
= =
−
= ⋅ − ⋅ = −
1 1 3 32 2seccos 42 2 22 2
3
θθ
⋅= = = − = −
−
1 1 4 2cot 2 2
tan 2 224
θθ
= = = − ⋅ = −
−
55. 3
tan , sin 0, so is in quadrant III4
θ θ θ= <
Solve for secθ : 2 2
2
sec 1 tan
sec 1 tan
θ θ
θ θ
= +
= ± +
Since is in quadrant III, sec 0θ θ < .
2
2
sec 1 tan
3 9 25 51 1
4 16 16 4
θ θ= − +
= − + = − + = − = −
1 4cos
sec 5θ
θ= = −
2
2
sin 1 cos
4 16 9 31 1
5 25 25 5
θ θ= − −
= − − − = − − = − = −
1 1 5csc
3sin 35
θθ
= = = − −
1 1 4cot
3tan 34
θθ
= = =
56. 4
cot , cos 0, so is in quadrant III3
θ θ θ= <
1 1 3tan
4cot 43
θθ
= = =
Solve for secθ : 2 2
2
sec 1 tan
sec 1 tan
θ θ
θ θ
= +
= ± +
Since is in quadrant III, sec 0θ θ < .
2
2
sec 1 tan
3 9 25 51 1
4 16 16 4
θ θ= − +
= − + = − + = − = −
1 4cos
sec 5θ
θ= = −
Section 6.3: Properties of the Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
2
2
sin 1 cos
4 16 9 31 1
5 25 25 5
θ θ= − −
= − − − = − − = − = −
1 1 5csc
3sin 35
θθ
= = = − −
57. 1
tan , sin 0, so is in quadrant II3
θ θ θ= − >
Solve for secθ : 2 2
2
sec 1 tan
sec 1 tan
θ θ
θ θ
= +
= ± +
Since is in quadrant II, sec 0θ θ < .
2
2
sec 1 tan
1 1 10 101 1
3 9 9 3
θ θ= − +
= − + − = − + = − = −
1 1 3 3 10cos
sec 1010103
θθ
= = = − = −
−
2
2
sin 1 cos
3 10 901 1
10 100
10 10
100 10
θ θ= −
= − − = −
= =
1 1csc 10
sin 1010
θθ
= = =
1 1cot 3
1tan3
θθ
= = = − −
58. sec 2, tan 0, so is in quadrant IIIθ θ θ= − >
Solve for tanθ : 2 2
2
sec 1 tan
tan sec 1
θ θ
θ θ
= +
= ± −
2 2tan sec 1 ( 2) 1 4 1 3θ θ= − = − − = − =
1 1cos
sec 2θ
θ= = −
1 1 3 3cot
tan 33 3θ
θ= = ⋅ =
2
2
sin 1 cos
1 1 3 31 1
2 4 4 2
θ θ= − −
= − − − = − − = − = −
1 1 2 3 2 3csc
sin 33 332
θθ
= = = − ⋅ = −
−
59. 3
sin( 60º ) sin 60º2
− = − = −
60. 3
cos( 30º ) cos30º2
− = =
61. 3
tan( 30º ) tan 30º3
− = − = −
62. 2
sin( 135º ) sin135º2
− = − = −
63. sec( 60º ) sec60º 2− = =
64. csc( 30º ) csc30º 2− = − = −
65. sin( 90º ) sin 90º 1− = − = −
66. cos( 270º ) cos 270º 0− = =
67. tan tan 14 4
π π − = − = −
68. sin( ) sin 0−π = − π =
69. 2
cos cos4 4 2
π π − = =
70. 3
sin sin3 3 2
π π − = − = −
71. tan( ) tan 0−π = − π =
72. 3 3
sin sin ( 1) 12 2
π π − = − = − − =
73. csc csc 24 4
π π − = − = −
74. ( )sec sec 1−π = π = −
Chapter 6: Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
75. 2 3
sec sec6 6 3
π π − = =
76. 2 3
csc csc3 3 3
π π − = − = −
77. ( ) ( )2 2sin 40º cos 40º 1+ =
78. ( ) ( )2 2sec 18º tan 18º 1− =
79. ( ) ( ) ( ) ( )1
sin 80º csc 80º sin 80º 1sin 80º
= ⋅ =
80. ( ) ( ) ( ) ( )1
tan 10º cot 10º tan 10º 1tan 10º
= ⋅ =
81. ( ) ( )( ) ( ) ( )sin 40º
tan 40º tan 40º tan 40º 0cos 40º
− = − =
82. ( ) ( )( ) ( ) ( )cos 20º
cot 20º cot 20º cot 20º 0sin 20º
− = − =
83. ( ) ( ) ( ) ( )( ) ( )
( ) ( )
cos 400º sec 40º cos 40º 360º sec 40º
cos 40º sec 40º
1cos 40º 1
cos 40º
⋅ = + ⋅
= ⋅
= ⋅ =
84. ( ) ( ) ( ) ( )( ) ( )
( ) ( )
tan 200º cot 20º tan 20º 180º cot 20º
tan 20º cot 20º
1tan 20º 1
tan 20º
⋅ = + ⋅
= ⋅
= ⋅ =
85. 25 25
sin csc sin csc12 12 12 12
π π π π − = −
24
sin csc12 12 12
sin csc 212 12
sin csc12 12
1sin 1
12sin
12
π π π
π π π
π π
ππ
= − + = − + = − = − ⋅ = −
86. 37 37
sec cos sec cos18 18 18 18
π π π π − =
36sec cos
18 18 18
sec cos 218 18
sec cos18 18
1sec 1
18sec
18
π π π
π π π
π π
ππ
= + = + = = ⋅ =
87. ( )( ) ( )
( )( ) ( )
( )( ) ( )
( ) ( )
sin 20ºtan 200º
cos 380º
sin 20ºtan 20º 180º
cos 20º 360º
sin 20ºtan 20º
cos 20º
tan 20º tan 20º 0
−+
−= + +
+
−= +
= − + =
88. ( )
( ) ( )
( )( ) ( )
sin 70ºtan 70º
cos 430º
sin 70ºtan 70º
cos 430º
+ −−
= −
( )( ) ( )
( )( ) ( )
( ) ( )
sin 70ºtan 70º
cos 70º 360º
sin 70ºtan 70º
cos 70º
tan 70º tan 70º 0
= −+
= −
= − =
89. If sin 0.3θ = , then
( ) ( )sin sin 2 sin 4
0.3 0.3 0.3 0.9
θ θ θ+ + π + + π= + + =
90. If cos 0.2θ = , then
( ) ( )cos cos 2 cos 4
0.2 0.2 0.2 0.6
θ θ θ+ + π + + π= − + + =
91. If tan 3θ = , then
( ) ( )tan tan tan 2
3 3 3 9
θ θ θ+ + π + + π= + + =
Section 6.3: Properties of the Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
92. If cot 2θ = − , then
( ) ( )( ) ( )
cot cot cot 2
2 2 2 6
θ θ θ+ − π + − π
= − + − + − = −
93. sin1º sin 2º sin 3º ... sin 357º
sin 358º sin 359º
sin1º sin 2º sin 3º sin(360º 3º )
sin(360º 2º ) sin(360º 1º )
sin1º sin 2º sin 3º sin( 3
+ + + ++ +
= + + + ⋅⋅⋅ + −+ − + −
= + + + ⋅⋅⋅ + −
( )
º )
sin( 2º ) sin( 1º )
sin1º sin 2º sin 3º sin 3º sin 2º sin1º
sin 180º 0
+ − + −= + + + ⋅⋅⋅ − − −= =
94. cos1º cos 2º cos3º cos357º
cos358º cos359º
+ + + ⋅⋅⋅ ++ +
cos1º cos 2º cos3º ... cos(360º 3º )
cos(360º 2º ) cos(360º 1º )
cos1º cos 2º cos3º ... cos( 3º )
cos( 2º ) cos( 1º )
cos1º cos 2º cos3º ... cos3º
= + + + + −+ − + −
= + + + + −+ − + −
= + + + + cos 2º cos1º
2cos1º 2cos 2º 2cos3º ... 2cos178º
2cos179º cos180º
2cos1º 2cos 2º 2cos3º ... 2cos(180º 2º )
+ += + + + +
+ += + + + + −
( ) 2cos(180º 1º ) cos 180º
2cos1º 2cos 2º 2cos3º ... 2cos 2º
2cos1º cos180º
cos180º 1
+ − += + + + −
− += = −
95. The domain of the sine function is the set of all real numbers.
96. The domain of the cosine function is the set of all real numbers.
97. ( ) tanf θ θ= is not defined for numbers that are
odd multiples of 2
π.
98. ( ) cotf θ θ= is not defined for numbers that are
multiples of π .
99. ( ) secf θ θ= is not defined for numbers that are
odd multiples of 2
π.
100. ( ) cscf θ θ= is not defined for numbers that are
multiples of π .
101. The range of the sine function is the set of all real numbers between 1− and 1, inclusive.
102. The range of the cosine function is the set of all real numbers between 1− and 1, inclusive.
103. The range of the tangent function is the set of all real numbers.
104. The range of the cotangent function is the set of all real numbers.
105. The range of the secant function is the set of all real numbers greater than or equal to 1 and all real numbers less than or equal to 1− .
106. The range of the cosecant function is the set of all real number greater than or equal to 1 and all real numbers less than or equal to 1− .
107. The sine function is odd because sin( ) sinθ θ− = − . Its graph is symmetric with
respect to the origin.
108. The cosine function is even because cos( ) cosθ θ− = . Its graph is symmetric with
respect to the y-axis.
109. The tangent function is odd because tan( ) tanθ θ− = − . Its graph is symmetric with
respect to the origin.
110. The cotangent function is odd because cot( ) cotθ θ− = − . Its graph is symmetric with
respect to the origin.
111. The secant function is even because sec( ) secθ θ− = . Its graph is symmetric with
respect to the y-axis.
112. The cosecant function is odd because csc( ) cscθ θ− = − . Its graph is symmetric with
respect to the origin.
113. a. 1
( ) ( )3
f a f a− = − = −
b. ( ) ( 2 ) ( 4 )
( ) ( ) ( )
1 1 11
3 3 3
f a f a f a
f a f a f a
+ + π + + π= + +
= + + =
Chapter 6: Trigonometric Functions
582
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
114. a. 1
( ) ( )4
f a f a− = =
b. ( ) ( 2 ) ( 2 )
( ) ( ) ( )
1 1 1
4 4 43
4
f a f a f a
f a f a f a
+ + π + − π= + +
= + +
=
115. a. ( ) ( ) 2f a f a− = − = −
b. ( ) ( ) ( 2 )
( ) ( ) ( )
2 2 2 6
f a f a f a
f a f a f a
+ + π + + π= + += + + =
116. a. ( ) ( ) ( 3) 3f a f a− = − = − − =
b. ( ) ( ) ( 4 )
( ) ( ) ( )
3 ( 3) ( 3)
9
f a f a f a
f a f a f a
+ + π + + π= + += − + − + −= −
117. a. ( ) ( ) 4f a f a− = = −
b. ( ) ( 2 ) ( 4 )
( ) ( ) ( )
4 ( 4) ( 4)
12
f a f a f a
f a f a f a
+ + π + + π= + += − + − + −= −
118. a. ( ) ( ) 2f a f a− = − = −
b. ( ) ( 2 ) ( 4 )
( ) ( ) ( )
2 2 2
6
f a f a f a
f a f a f a
+ + π + + π= + += + +=
119. Since 500 1
tan1500 3
y
xθ = = = , then
2 2 2 9 1 10
10
r x y
r
= + = + =
=
1 1sin
1 9 10θ = =
+.
5 55
1 133 10
5 5 5 10
5 10 15.8 minutes
T = − + ⋅
= − +
= ≈
120. a. 1
tan4
y
xθ = = for 0
2θ π< < .
2 2 2 16 1 17
17
r x y
r
= + = + =
=
Thus, 1
sin17
θ = .
2 1( ) 1
11 43417
2 17 2 171 1 2.75 hours
3 3
T θ = + − ⋅⋅
= + − = ≈
b. Since 1
tan4
θ = , 4x = . Sally heads
directly across the sand to the bridge, crosses the bridge, and heads directly across the sand to the other house.
c. θ must be larger than 14º , or the road will
not be reached and she cannot get across the river.
121. Let ( , )P x y= be the point on the unit circle that
corresponds to an angle t. Consider the equation
tany
t ax
= = . Then y ax= . Now 2 2 1x y+ = ,
so 2 2 2 1x a x+ = . Thus, 2
1
1x
a= ±
+ and
21
ay
a= ±
+. That is, for any real number a ,
there is a point ( , )P x y= on the unit circle for
which tan t a= . In other words, tan t−∞ < < ∞ , and the range of the tangent
function is the set of all real numbers.
122. Let ( , )P x y= be the point on the unit circle that
corresponds to an angle t. Consider the equation
cotx
t ay
= = . Then x ay= . Now 2 2 1x y+ = ,
so 2 2 2 1a y y+ = . Thus, 2
1
1y
a= ±
+ and
21
ax
a= ±
+. That is, for any real number a ,
there is a point ( , )P x y= on the unit circle for
which cot t a= . In other words, cot t−∞ < < ∞ , and the range of the tangent function is the set of all real numbers.
Section 6.3: Properties of the Trigonometric Functions
583
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
123. Suppose there is a number p, 0 2p< < π for
which sin( ) sinpθ θ+ = for all θ . If 0θ = ,
then ( )sin 0 sin sin 0 0p p+ = = = ; so that
p = π . If 2
πθ = then sin sin2 2
pπ π + =
.
But p = π . Thus, 3
sin 1 sin 12 2
π π = − = =
,
or 1 1− = . This is impossible. The smallest positive number p for which sin( ) sinpθ θ+ =
for all θ must then be 2p = π .
124. Suppose there is a number p, 0 2p< < π , for
which cos( ) cospθ θ+ = for all θ . If 2
θ π= ,
then cos cos 02 2
pπ π + = =
; so that p = π .
If 0θ = , then ( ) ( )cos 0 cos 0p+ = . But
p = π . Thus ( ) ( )cos 1 cos 0 1π = − = = , or
1 1− = . This is impossible. The smallest positive number p for which cos( ) cospθ θ+ =
for all θ must then be 2p = π .
125. 1
seccos
θθ
= : Since cosθ has period 2π , so
does sec .θ
126. 1
cscsin
θθ
= : Since sinθ has period 2π , so
does csc .θ
127. If ( , )P a b= is the point on the unit circle
corresponding to θ , then ( , )Q a b= − − is the
point on the unit circle corresponding to θ + π .
Thus, tan( ) tanb b
a aθ θ−+ π = = =
−. If there
exists a number , 0 ,p p π< < for which
tan( ) tanpθ θ+ = for all θ , then if 0θ = ,
( ) ( )tan tan 0 0.p = = But this means that p is a
multiple of π . Since no multiple of π exists in
the interval ( )0,π , this is impossible. Therefore,
the fundamental period of ( ) tanf θ θ= is π .
128.1
cot = tan
θθ
: Since tanθ has period π , so does
cotθ .
129. Let ( , )P a b= be the point on the unit circle
corresponding to θ . Then 1 1
cscsinb
θθ
= =
1 1sec
cosaθ
θ= =
1 1cot
tan
a
bba
θθ
= = =
130. Let ( , )P a b= be the point on the unit circle
corresponding to θ . Then sin
tancos
b
a
θθθ
= =
coscot
sin
a
b
θθθ
= =
131. 2 2 2
2 2 2 2 2
2 2 2 2
2 2
(sin cos ) (sin sin ) cos
sin cos sin sin cos
sin (cos sin ) cos
sin cos
1
θ φ θ φ θθ φ θ φ θθ φ φ θθ θ
+ += + +
= + += +=
132 – 136. Answers will vary.
Chapter 6: Trigonometric Functions
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Section 6.4
1. 23y x=
Using the graph of 2y x= , vertically stretch the
graph by a factor of 3.
2. 2y x=
Using the graph of y x= , compress
horizontally by a factor of 1
2.
3. 1; 2
π
4. 3; π
5. 3; 2
6 3
π π=
6. True
7. False; The period is 2
2π
π=
8. True
9. a. The graph of siny x= crosses the y-axis at
the point (0, 0), so the y-intercept is 0.
b. The graph of siny x= is increasing for
2 2x
π π− < < .
c. The largest value of siny x= is 1.
d. sin 0 when 0, , 2x x= = π π .
e. 3
sin 1 when , ;2 2
3sin 1 when , .
2 2
x x
x x
π π= = −
π π= − = −
f. 1 5 7 11
sin when , , ,2 6 6 6 6
x xπ π π π= − = − −
g. The x-intercepts of sin x are
{ }| , k an integerx x kπ=
10. a. The graph of cosy x= crosses the y-axis at
the point (0, 1), so the y-intercept is 1. b. The graph of cosy x= is decreasing for
0 x< < π . c. The smallest value of cosy x= is 1− .
d. 3
cos 0 when ,2 2
x xπ π= =
e. cos 1 when 2 , 0, 2 ;
cos 1 when , .
x x
x x
= = − π π= − = −π π
f. 3 11 11
cos when , , ,2 6 6 6 6
x xπ π π π= = − −
g. The x-intercepts of cos x are
( )2 1| , k an integer
2
kx x
π+ =
11. 2siny x=
This is in the form sin( )y A xω= where 2A =
and 1ω = . Thus, the amplitude is 2 2A = =
and the period is 2 2
21
Tωπ π= = = π .
12. 3cosy x=
This is in the form cos( )y A xω= where 3A =
and 1ω = . Thus, the amplitude is 3 3A = =
and the period is 2 2
21
Tωπ π= = = π .
13. 4cos(2 )y x= −
This is in the form cos( )y A xω= where
4A = − and 2ω = . Thus, the amplitude is
4 4A = − = and the period is
2 2
2T
ωπ π= = = π .
Section 6.4: Graphs of the Sine and Cosine Functions
585
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
14. 1
sin2
y x = −
This is in the form sin( )y A xω= where 1A = −
and 1
2ω = . Thus, the amplitude is 1 1A = − =
and the period is 12
2 24T
ωπ π= = = π .
15. 6sin( )y x= π
This is in the form sin( )y A xω= where 6A =
and ω π= . Thus, the amplitude is 6 6A = =
and the period is 2 2
2Tωπ π= = =
π.
16. 3cos(3 )y x= −
This is in the form cos( )y A xω= where 3A = −
and 3ω = . Thus, the amplitude is 3 3A = − =
and the period is 2 2
3T
ωπ π= = .
17. 1 3
cos2 2
y x = −
This is in the form cos( )y A xω= where
1
2A = − and
3
2ω = . Thus, the amplitude is
1 1
2 2A = − = and the period is
32
2 2 4
3T
ωπ π π= = = .
18. 4 2
sin3 3
y x =
This is in the form sin( )y A xω= where 4
3A =
and 2
3ω = . Thus, the amplitude is
4 4
3 3A = =
and the period is 23
2 23T
ωπ π= = = π .
19. 5 2 5 2
sin sin3 3 3 3
y x xπ π = − = −
This is in the form sin( )y A xω= where 5
3A = −
and 2
3
πω = . Thus, the amplitude is
5 5
3 3A = − = and the period is
2
3
2 23T
πωπ π= = = .
20. 9 3 9 3
cos cos5 2 5 2
y x xπ π = − =
This is in the form cos( )y A xω= where 9
5A =
and 3
2
πω = . Thus, the amplitude is
9 9
5 5A = = and the period is
3
2
2 2 4
3T
πωπ π= = = .
21. F
22. E
23. A
24. I
25. H
26. B
27. C
28. G
29. J
30. D
31. A
32. C
33. B
34. D
35. Comparing 4cosy x= to ( )cosy A xω= , we
find 4A = and 1ω = . Therefore, the amplitude
is 4 4= and the period is 2
21
π π= . Because
the amplitude is 4, the graph of 4cosy x= will
lie between 4− and 4 on the y-axis. Because the
Chapter 6: Trigonometric Functions
586
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
period is 2π , one cycle will begin at 0x = and
end at 2x π= . We divide the interval [ ]0,2π
into four subintervals, each of length 2
4 2
π π= by
finding the following values:
0, 2
π, π ,
3
2
π, and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
4cosy x= , we multiply the y-coordinates of the
five key points for cosy x= by 4A = . The five
key points are
( )0,4 , ,02
π
, ( ), 4π − , 3
,02
π
, ( )2 ,4π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]4,4− .
36. Comparing 3siny x= to ( )siny A xω= , we
find 3A = and 1ω = . Therefore, the amplitude
is 3 3= and the period is 2
21
π π= . Because
the amplitude is 3, the graph of 3siny x= will
lie between 3− and 3 on the y-axis. Because the period is 2π , one cycle will begin at 0x = and
end at 2x π= . We divide the interval [ ]0,2π
into four subintervals, each of length 2
4 2
π π= by
finding the following values:
0, 2
π, π ,
3
2
π, and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-
coordinates of the five key points for 3siny x= ,
we multiply the y-coordinates of the five key points for siny x= by 3A = . The five key
points are ( )0,0 , ,32
π
, ( ),0π , 3
, 32
π −
,
( )2 ,0π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]3,3− .
37. Comparing 4siny x= − to ( )siny A xω= , we
find 4A = − and 1ω = . Therefore, the amplitude
is 4 4− = and the period is 2
21
π π= . Because
the amplitude is 4, the graph of 4siny x= − will
lie between 4− and 4 on the y-axis. Because the period is 2π , one cycle will begin at 0x = and
end at 2x π= . We divide the interval [ ]0,2π
into four subintervals, each of length 2
4 2
π π= by
finding the following values: 0, 2
π, π ,
3
2
π, 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
4siny x= − , we multiply the y-coordinates of
the five key points for siny x= by 4A = − . The
five key points are
( )0,0 , , 42
π −
, ( ),0π , 3
, 42
π
, ( )2 ,0π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
Section 6.4: Graphs of the Sine and Cosine Functions
587
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]4,4− .
38. Comparing 3cosy x= − to ( )cosy A xω= , we
find 3A = − and 1ω = . Therefore, the amplitude
is 3 3− = and the period is 2
21
π π= . Because
the amplitude is 3, the graph of 3cosy x= − will
lie between 3− and 3 on the y-axis. Because the period is 2π , one cycle will begin at 0x = and
end at 2x π= . We divide the interval [ ]0,2π
into four subintervals, each of length 2
4 2
π π= by
finding the following values:
0, 2
π, π ,
3
2
π, and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
3cosy x= − , we multiply the y-coordinates of
the five key points for cosy x= by 3A = − . The
five key points are
( )0, 3− , ,02
π
, ( ),3π , 3
,02
π
, ( )2 , 3π −
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
�� ��2� 2�
5(�, 3)(��, 3)
(0, �3)
�5
y
x
(2�, �3)
3�–––2(
, 0)
, 0)
�––2(
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]3,3− .
39. Comparing ( )cos 4y x= to ( )cosy A xω= , we
find 1A = and 4ω = . Therefore, the amplitude
is 1 1= and the period is 2
4 2
π π= . Because the
amplitude is 1, the graph of ( )cos 4y x= will lie
between 1− and 1 on the y-axis. Because the
period is2
π, one cycle will begin at 0x = and
end at 2
xπ= . We divide the interval 0,
2
π
into four subintervals, each of length / 2
4 8
π π=
by finding the following values:
0, 8
π,
4
π,
3
8
π, and
2
π
These values of x determine the x-coordinates of the five key points on the graph. The five key points are
( )0,1 , ,08
π
, , 14
π −
, 3
,08
π
, ,12
π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,1− .
40. Comparing ( )sin 3y x= to ( )siny A xω= , we
find 1A = and 3ω = . Therefore, the amplitude
is 1 1= and the period is 2
3
π. Because the
amplitude is 1, the graph of ( )sin 3y x= will lie
Chapter 6: Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
between 1− and 1 on the y-axis. Because the
period is2
3
π, one cycle will begin at 0x = and
end at 2
3x
π= . We divide the interval 2
0,3
π
into four subintervals, each of length 2 / 3
4 6
π π=
by finding the following values:
0, 6
π,
3
π,
2
π, and
2
3
π
These values of x determine the x-coordinates of the five key points on the graph. The five key points are
( )0,0 , ,16
π
, ,03
π
, , 12
π −
, 2
,03
π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,1− .
41. Since sine is an odd function, we can plot the
equivalent form ( )sin 2y x= − .
Comparing ( )sin 2y x= − to ( )siny A xω= , we
find 1A = − and 2ω = . Therefore, the
amplitude is 1 1− = and the period is 2
2
π π= .
Because the amplitude is 1, the graph of
( )sin 2y x= − will lie between 1− and 1 on the
y-axis. Because the period isπ , one cycle will begin at 0x = and end at x π= . We divide the
interval [ ]0,π into four subintervals, each of
length 4
π by finding the following values:
0, 4
π,
2
π,
3
4
π, and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
( )sin 2y x= − , we multiply the y-coordinates of
the five key points for siny x= by 1A = − .The
five key points are
( )0,0 , , 14
π −
, ,02
π
, 3
,14
π
, ( ),0π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
y
x
2
�2
2��2�
(�, 0)
��
3�–––4( , 1)��–––
4( , 1)
�––4( , �1)
�––2( , 0)(0, 0)
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,1− .
42. Since cosine is an even function, we can plot the
equivalent form ( )cos 2y x= .
Comparing ( )cos 2y x= to ( )cosy A xω= , we
find 1A = and 2ω = . Therefore, the amplitude
is 1 1= and the period is 2
2
π π= . Because the
amplitude is 1, the graph of ( )cos 2y x= will lie
between 1− and 1 on the y-axis. Because the period isπ , one cycle will begin at 0x = and
end at x π= . We divide the interval [ ]0,π into
four subintervals, each of length 4
π by finding
the following values:
0, 4
π,
2
π,
3
4
π, and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
( )cos 2y x= , we multiply the y-coordinates of
the five key points for cosy x= by 1A = .The
Section 6.4: Graphs of the Sine and Cosine Functions
589
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
five key points are
( )0,1 , ,04
π
, , 12
π −
, 3
,04
π
, ( ),1π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,1− .
43. Comparing 1
2sin2
y x =
to ( )siny A xω= ,
we find 2A = and 1
2ω = . Therefore, the
amplitude is 2 2= and the period is 2
41/ 2
π π= .
Because the amplitude is 2, the graph of 1
2sin2
y x =
will lie between 2− and 2 on the
y-axis. Because the period is 4π , one cycle will begin at 0x = and end at 4x π= . We divide
the interval [ ]0,4π into four subintervals, each
of length 4
4
π π= by finding the following
values: 0, π , 2π , 3π , and 4π These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
12sin
2y x
=
, we multiply the y-coordinates of
the five key points for siny x= by 2A = . The
five key points are
( )0,0 , ( ), 2π , ( )2 ,0π , ( )3 , 2π − , ( )4 ,0π
We plot these five points and fill in the graph of the curve. We then extend the graph in either
direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]2,2− .
44. Comparing 1
2cos4
y x =
to ( )cosy A xω= ,
we find 2A = and 1
4ω = . Therefore, the
amplitude is 2 2= and the period is 2
81/ 4
π π= .
Because the amplitude is 2, the graph of 1
2cos4
y x =
will lie between 2− and 2 on
the y-axis. Because the period is 8π , one cycle will begin at 0x = and end at 8x π= . We
divide the interval [ ]0,8π into four subintervals,
each of length 8
24
π π= by finding the following
values: 0, 2π , 4π , 6π , and 8π These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
12cos
4y x
=
, we multiply the y-coordinates
of the five key points for cosy x= by
2A = .The five key points are
( )0,2 , ( )2 ,0π , ( )4 , 2π − , ( )6 ,0π , ( )8 ,2π
We plot these five points and fill in the graph of the curve. We then extend the graph in either
Chapter 6: Trigonometric Functions
590
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
direction to obtain the graph shown below.
2
�2
�8� �4� 8�4�
y
x
(0, 2)
(2�, 0)(6�, 0)
(8�, 2)
(�8�, 2)
(�2�, 0)
(�4�, �2) (4�, �2)
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]2,2− .
45. Comparing ( )1cos 2
2y x= − to ( )cosy A xω= ,
we find 1
2A = − and 2ω = . Therefore, the
amplitude is 1 1
2 2− = and the period is
2
2
π π= .
Because the amplitude is 1
2, the graph of
( )1cos 2
2y x= − will lie between
1
2− and
1
2 on
the y-axis. Because the period isπ , one cycle will begin at 0x = and end at x π= . We divide
the interval [ ]0,π into four subintervals, each of
length 4
π by finding the following values:
0, 4
π,
2
π,
3
4
π, and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
( )1cos 2
2y x= − , we multiply the y-coordinates
of the five key points for cosy x= by
1
2A = − .The five key points are
10,
2 −
, ,04
π
, 1
,2 2
π
, 3
,04
π
, 1
,2
π −
We plot these five points and fill in the graph of the curve. We then extend the graph in either
direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is 1 1
,2 2
− .
46. Comparing 1
4sin8
y x = −
to ( )siny A xω= ,
we find 4A = − and 1
8ω = . Therefore, the
amplitude is 4 4− = and the period is
216
1/ 8
π π= . Because the amplitude is 4, the
graph of 1
4sin8
y x = −
will lie between 4−
and 4 on the y-axis. Because the period is16π , one cycle will begin at 0x = and end at
16x π= . We divide the interval [ ]0,16π into
four subintervals, each of length 16
44
π π= by
finding the following values: 0, 4π , 8π , 12π , and 16π These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
14sin
8y x
= −
, we multiply the y-coordinates
of the five key points for siny x= by 4A = − .
The five key points are
( )0,0 , ( )4 , 4π − , ( )8 ,0π , ( )12 ,4π , ( )16 ,0π
We plot these five points and fill in the graph of the curve. We then extend the graph in either
Section 6.4: Graphs of the Sine and Cosine Functions
591
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
direction to obtain the graph shown below.
�8��16�
5 (12�, 4)
(4�, �4)
(16�, 0)
(8�, 0)
(0, 0)
�5
y
x
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]4,4− .
47. We begin by considering 2siny x= . Comparing
2siny x= to ( )siny A xω= , we find 2A =
and 1ω = . Therefore, the amplitude is 2 2=
and the period is 2
21
π π= . Because the
amplitude is 2, the graph of 2siny x= will lie
between 2− and 2 on the y-axis. Because the period is 2π , one cycle will begin at 0x = and
end at 2x π= . We divide the interval [ ]0,2π
into four subintervals, each of length 2
4 2
π π= by
finding the following values:
0, 2
π, π ,
3
2
π, and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
2sin 3y x= + , we multiply the y-coordinates of
the five key points for siny x= by 2A = and
then add 3 units. Thus, the graph of 2sin 3y x= + will lie between 1 and 5 on the y-
axis. The five key points are
( )0,3 , ,52
π
, ( ),3π , 3
,12
π
, ( )2 ,3π
We plot these five points and fill in the graph of the curve. We then extend the graph in either
direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,5 .
48. We begin by considering 3cosy x= . Comparing
3cosy x= to ( )cosy A xω= , we find 3A =
and 1ω = . Therefore, the amplitude is 3 3=
and the period is 2
21
π π= . Because the
amplitude is 3, the graph of 3cosy x= will lie
between 3− and 3 on the y-axis. Because the period is 2π , one cycle will begin at 0x = and
end at 2x π= . We divide the interval [ ]0,2π
into four subintervals, each of length 2
4 2
π π= by
finding the following values:
0, 2
π, π ,
3
2
π, and 2π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
3cos 2y x= + , we multiply the y-coordinates of
the five key points for cosy x= by 3A = and
then add 2 units. Thus, the graph of 3cos 2y x= + will lie between 1− and 5 on the
y-axis. The five key points are
( )0,5 , , 22
π
, ( ), 1π − , 3
, 22
π
, ( )2 ,5π
We plot these five points and fill in the graph of the curve. We then extend the graph in either
Chapter 6: Trigonometric Functions
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direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,5− .
49. We begin by considering ( )5cosy xπ= .
Comparing ( )5cosy xπ= to ( )cosy A xω= , we
find 5A = and ω π= . Therefore, the amplitude
is 5 5= and the period is 2
2π
π= . Because the
amplitude is 5, the graph of ( )5cosy xπ= will
lie between 5− and 5 on the y-axis. Because the period is 2 , one cycle will begin at 0x = and
end at 2x = . We divide the interval [ ]0,2 into
four subintervals, each of length 2 1
4 2= by
finding the following values:
0, 1
2, 1,
3
2, and 2
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
( )5cos 3y xπ= − , we multiply the y-coordinates
of the five key points for cosy x= by 5A =
and then subtract 3 units. Thus, the graph of
( )5cos 3y xπ= − will lie between 8− and 2 on
the y-axis. The five key points are
( )0,2 , 1
, 32
−
, ( )1, 8− , 3
, 32
−
, ( )2, 2
We plot these five points and fill in the graph of the curve. We then extend the graph in either
direction to obtain the graph shown below.
�� � 2��2�
3
�9
y
x
3––2( , �3)
1––2( , �3)
3––2(� , �3)
(0, 2) (2, 2)
(1, �8)
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]8,2− .
50. We begin by considering 4sin2
y xπ =
.
Comparing 4sin2
y xπ =
to ( )siny A xω= ,
we find 4A = and 2
πω = . Therefore, the
amplitude is 4 4= and the period is 2
4/ 2
ππ
= .
Because the amplitude is 4, the graph of
4sin2
y xπ =
will lie between 4− and 4 on
the y-axis. Because the period is 4 , one cycle will begin at 0x = and end at 4x = . We divide
the interval [ ]0,4 into four subintervals, each of
length 4
14
= by finding the following values:
0, 1, 2, 3, and 4 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
4sin 22
y xπ = −
, we multiply the y-
coordinates of the five key points for siny x=
by 4A = and then subtract 2 units. Thus, the
graph of 4sin 22
y xπ = −
will lie between 6−
and 2 on the y-axis. The five key points are
( )0, 2− , ( )1,2 , ( )2, 2− , ( )3, 6− , ( )4, 2−
We plot these five points and fill in the graph of the curve. We then extend the graph in either
Section 6.4: Graphs of the Sine and Cosine Functions
593
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]6,2− .
51. We begin by considering 6sin3
y xπ = −
.
Comparing 6sin3
y xπ = −
to ( )siny A xω= ,
we find 6A = − and 3
πω = . Therefore, the
amplitude is 6 6− = and the period is 2
6/ 3
ππ
= .
Because the amplitude is 6, the graph of
6sin3
y xπ =
will lie between 6− and 6 on the
y-axis. Because the period is 6, one cycle will begin at 0x = and end at 6x = . We divide the
interval [ ]0,6 into four subintervals, each of
length 6 3
4 2= by finding the following values:
0, 3
2, 3,
9
2, and 6
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
6sin 43
y xπ = − +
, we multiply the y-
coordinates of the five key points for siny x=
by 6A = − and then add 4 units. Thus, the graph
of 6sin 43
y xπ = − +
will lie between 2− and
10 on the y-axis. The five key points are
( )0,4 , 3
, 22
−
, ( )3, 4 , 9
,102
, ( )6,4
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]2,10− .
52. We begin by considering 3cos4
y xπ = −
.
Comparing 3cos4
y xπ = −
to ( )cosy A xω= ,
we find 3A = − and 4
πω = . Therefore, the
amplitude is 3 3− = and the period is 2
8/ 4
ππ
= .
Because the amplitude is 3, the graph of
3cos4
y xπ = −
will lie between 3− and 3 on
the y-axis. Because the period is 8, one cycle will begin at 0x = and end at 8x = . We divide
the interval [ ]0,8 into four subintervals, each of
length 8
24
= by finding the following values:
0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
3cos 24
y xπ = − +
, we multiply the y-
coordinates of the five key points for cosy x=
by 3A = − and then add 2 units. Thus, the graph
Chapter 6: Trigonometric Functions
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of 3cos 24
y xπ = − +
will lie between 1− and
5 on the y-axis. The five key points are
( )0, 1− , ( )2, 2 , ( )4,5 , ( )6,2 , ( )8, 1−
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,5− .
53. ( ) ( )5 3sin 2 3sin 2 5y x x= − = − +
We begin by considering ( )3sin 2y x= − .
Comparing ( )3sin 2y x= − to ( )siny A xω= ,
we find 3A = − and 2ω = . Therefore, the
amplitude is 3 3− = and the period is 2
2
π π= .
Because the amplitude is 3, the graph of
( )3sin 2y x= − will lie between 3− and 3 on the
y-axis. Because the period is π , one cycle will begin at 0x = and end at x π= . We divide the
interval [ ]0,π into four subintervals, each of
length 4
π by finding the following values:
0, 4
π,
2
π,
3
4
π, and π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
( )3sin 2 5y x= − + , we multiply the y-
coordinates of the five key points for siny x=
by 3A = − and then add 5 units. Thus, the graph
of ( )3sin 2 5y x= − + will lie between 2 and 8
on the y-axis. The five key points are
( )0,5 , , 24
π
, ,52
π
, 3
,84
π
, ( ),5π
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
y
10
π−π
( ),5π( )0,5
,24
π
3,8
4
π
,52
π
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]2,8 .
54. ( ) ( )2 4cos 3 4cos 3 2y x x= − = − +
We begin by considering ( )4cos 3y x= − .
Comparing ( )4cos 3y x= − to ( )cosy A xω= ,
we find 4A = − and 3ω = . Therefore, the
amplitude is 4 4− = and the period is 2
3
π.
Because the amplitude is 4, the graph of
( )4cos 3y x= − will lie between 4− and 4 on
the y-axis. Because the period is 2
3
π, one cycle
will begin at 0x = and end at 2
3x
π= . We
divide the interval 2
0,3
π
into four
subintervals, each of length 2 / 3
4 6
π π= by
finding the following values:
0, 6
π,
3
π,
2
π, and
2
3
π
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
( )4cos 3 2y x= − + , we multiply the y-
Section 6.4: Graphs of the Sine and Cosine Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
coordinates of the five key points for cosy x=
by 4A = − and then adding 2 units. Thus, the
graph of ( )4cos 3 2y x= − + will lie between 2−
and 6 on the y-axis. The five key points are
( )0, 2− , , 26
π
, ,63
π
, , 22
π
, 2
, 23
π −
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
33
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]2,6− .
55. Since sine is an odd function, we can plot the
equivalent form 5 2
sin3 3
y xπ = −
.
Comparing 5 2
sin3 3
y xπ = −
to
( )siny A xω= , we find 5
3A = − and
2
3
πω = .
Therefore, the amplitude is 5 5
3 3− = and the
period is 2
32 / 3
ππ
= . Because the amplitude is
5
3, the graph of
5 2sin
3 3y x
π = −
will lie
between 5
3− and
5
3 on the y-axis. Because the
period is 3 , one cycle will begin at 0x = and
end at 3x = . We divide the interval [ ]0,3 into
four subintervals, each of length 3
4 by finding
the following values:
0, 3
4,
3
2,
9
4, and 3
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
5 2sin
3 3y x
π = −
, we multiply the y-
coordinates of the five key points for siny x=
by 5
3A = − .The five key points are
( )0,0 , 3 5
,4 3
−
, 3
,02
, 9 5
,4 3
, ( )3,0
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is 5 5
,3 3
− .
56. Since cosine is an even function, we consider the
equivalent form 9 3
cos5 2
y xπ =
. Comparing
9 3cos
5 2y x
π =
to ( )cosy A xω= , we find
9
5A = and
3
2
πω = . Therefore, the amplitude is
9 9
5 5= and the period is
2 4
3 / 2 3
ππ
= . Because
the amplitude is 9
5, the graph of
9 3cos
5 2y x
π =
will lie between 9
5− and
9
5
Chapter 6: Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
on the y-axis. Because the period is 4
3, one
cycle will begin at 0x = and end at 4
3x = . We
divide the interval 4
0,3
into four subintervals,
each of length 4 / 3 1
4 3= by finding the following
values:
0, 1
3,
2
3, 1 , and
4
3
These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
9 3cos
5 2y x
π =
, we multiply the y-coordinates
of the five key points for cosy x= by 9
5A = .
Thus, the graph of 9 3
cos5 2
y xπ = −
will lie
between 9
5− and
9
5 on the y-axis. The five key
points are 9
0,5
, 1
,03
, 2 9
,3 5
−
, ( )1,0 , 4 9
,3 5
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is 9 9
,5 5
− .
57. We begin by considering 3
cos2 4
y xπ = −
.
Comparing 3
cos2 4
y xπ = −
to
( )cosy A xω= , we find 3
2A = − and
4
πω = .
Therefore, the amplitude is 3 3
2 2− = and the
period is 2
8/ 4
ππ
= . Because the amplitude is 3
2,
the graph of 3
cos2 4
y xπ = −
will lie between
3
2− and
3
2 on the y-axis. Because the period is
8, one cycle will begin at 0x = and end at
8x = . We divide the interval [ ]0,8 into four
subintervals, each of length 8
24
= by finding the
following values: 0, 2, 4, 6, and 8 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
3 1cos
2 4 2y x
π = − +
, we multiply the y-
coordinates of the five key points for cosy x=
by 3
2A = − and then add
1
2 unit. Thus, the
graph of 3 1
cos2 4 2
y xπ = − +
will lie between
1− and 2 on the y-axis. The five key points are
( )0, 1− , 1
2,2
, ( )4, 2 , 1
6,2
, ( )8, 1−
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
2
�2
�8 �4 84
y
x
(0, �1) (8, �1)(�8, �1)
(6, )
(4, 2)(�4, 2)
1––2
(2, )1––2
Section 6.4: Graphs of the Sine and Cosine Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,2− .
58. We begin by considering 1
sin2 8
y xπ = −
.
Comparing 1
sin2 8
y xπ = −
to ( )siny A xω= ,
we find 1
2A = − and
8
πω = . Therefore, the
amplitude is 1 1
2 2− = and the period is
216
/ 8
ππ
= . Because the amplitude is 1
2, the
graph of 1
sin2 8
y xπ = −
will lie between 1
2−
and 1
2 on the y-axis. Because the period is 16,
one cycle will begin at 0x = and end at 16x = .
We divide the interval [ ]0,16 into four
subintervals, each of length 16
44
= by finding
the following values: 0, 4, 8, 12, and 16 These values of x determine the x-coordinates of the five key points on the graph. To obtain the y-coordinates of the five key points for
1 3sin
2 8 2y x
π = − +
, we multiply the y-
coordinates of the five key points for siny x=
by 1
2A = − and then add
3
2 units. Thus, the
graph of 1 3
sin2 8 2
y xπ = − +
will lie between
1 and 2 on the y-axis. The five key points are 3
0,2
, ( )4,1 , 3
8,2
, ( )12, 2 , 3
16,2
We plot these five points and fill in the graph of the curve. We then extend the graph in either direction to obtain the graph shown below.
�4�8�12�16 4 8 12 16�0.5
2.5y
x
(16, )
(12, 2)(�4, 2)
(4, 1)3––2
(8, )3––2
(0, )3––2
From the graph we can determine that the domain is all real numbers, ( ),−∞ ∞ and the
range is [ ]1,2 .
59. 2 2
3; ; 2A TT
ω π π= = π = = =π
3sin(2 )y x= ±
60. 2 2 1
2; 4 ;4 2
A TT
ω π π= = π = = =π
12sin
2y x
= ±
61. 2 2
3; 2;2
A TT
ω π π= = = = = π
3sin( )y x= ± π
62. 2 2
4; 1; 21
A TT
ω π π= = = = = π
4sin(2 )y x= ± π
63. The graph is a cosine graph with amplitude 5 and period 8.
Find ω : 2
8
8 2
2
8 4
ωω
ω
π=
= ππ π= =
The equation is: 5cos4
y xπ =
.
64. The graph is a sine graph with amplitude 4 and period 8π.
Chapter 6: Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Find ω : 2
8
8 2
2 1
8 4
ωω
ω
ππ =
π = ππ= =π
The equation is: 1
4sin4
y x =
.
65. The graph is a reflected cosine graph with amplitude 3 and period 4π.
Find ω : 2
4
4 2
2 1
4 2
ωω
ω
ππ =
π = ππ= =π
The equation is: 1
3cos2
y x = −
.
66. The graph is a reflected sine graph with amplitude 2 and period 4.
Find ω : 2
4
4 2
2
4 2
ωω
ω
π=
= ππ π= =
The equation is: 2sin2
y xπ = −
.
67. The graph is a sine graph with amplitude 3
4 and
period 1.
Find ω : 2
1
2ω
ω
π=
= π
The equation is: ( )3sin 2
4y x= π .
68. The graph is a reflected cosine graph with
amplitude 5
2 and period 2.
Find ω : 2
2
2 2
2
2
ωω
ω
π=
= ππ= = π
The equation is: ( )5cos
2y x= − π .
69. The graph is a reflected sine graph with
amplitude 1 and period 4
3
π.
Find ω : 4 2
34 6
6 3
4 2
ωω
ω
π π=
π = ππ= =π
The equation is: 3
sin2
y x = −
.
70. The graph is a reflected cosine graph with amplitude π and period 2π.
Find ω : 2
2
2 2
21
2
ωω
ω
ππ =
π = ππ= =π
The equation is: cosy x= −π .
71. The graph is a reflected cosine graph, shifted up
1 unit, with amplitude 1 and period 3
2.
Find ω : 3 2
23 4
4
3
ωω
ω
π=
= ππ=
The equation is: 4
cos 13
y xπ = − +
.
72. The graph is a reflected sine graph, shifted down
1 unit, with amplitude 1
2 and period
4
3
π.
Find ω : 4 2
34 6
6 3
4 2
ωω
ω
π π=
π = ππ= =π
The equation is: 1 3
sin 12 2
y x = − −
.
73. The graph is a sine graph with amplitude 3 and period 4.
Section 6.4: Graphs of the Sine and Cosine Functions
599
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Find ω : 2
4
4 2
2
4 2
ωω
ω
π=
= ππ π= =
The equation is: 3sin2
y xπ =
.
74. The graph is a reflected cosine graph with amplitude 2 and period 2.
Find ω : 2
2
2 2
2
2
ωω
ω
π=
= ππ= = π
The equation is: 2cos( )y x= − π .
75. The graph is a reflected cosine graph with
amplitude 4 and period 2
3
π.
Find ω : 2 2
32 6
63
2
ωω
ω
π π=
π = ππ= =π
The equation is: ( )4cos 3y x= − .
76. The graph is a sine graph with amplitude 4 and period π.
Find ω : 2
2
22
ωω
ω
ππ =
π = ππ= =
π
The equation is: ( )4sin 2y x= .
77. ( ) ( ) ( ) ( )/ 2 0 sin / 2 sin 0
/ 2 0 / 21 0 2
/ 2
f fπ ππ π
π π
− −=
−−= =
The average rate of change is 2
π.
78. ( ) ( ) ( ) ( )/ 2 0 cos / 2 cos 0
/ 2 0 / 20 1 2
/ 2
f fπ ππ π
π π
− −=
−−= = −
The average rate of change is 2
π− .
79. ( ) ( )
( ) ( )
1 1sin sin 0
/ 2 0 2 2 2/ 2 0 / 2
sin / 4 sin 0
/ 22
2 2 22/ 2 2
f fπ
ππ π
ππ
π π π
⋅ − ⋅ − =−
−=
= = ⋅ =
The average rate of change is 2
π.
80. ( ) cos 2 cos(2 0)
/ 2 (0) 2/ 2 0 / 2
cos( ) cos(0) 1 1
/ 2 / 22 4
2
f fπ
ππ π
ππ π
π π
⋅ − ⋅ − =−
− − −= =
= − ⋅ = −
The average rate of change is 4
π− .
81. ( )( ) ( )sin 4f g x x=
( )( ) ( )4 sin 4sing f x x x= =
Chapter 6: Trigonometric Functions
600
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
82. ( )( ) 1cos
2f g x x
=
( )( ) ( )1 1cos cos
2 2g f x x x= =
83. ( ) ( ) ( )2 cos 2cosf g x x x= − = −
( )( ) ( )cos 2g f x x= −
84. ( )( ) ( )3 sin 3sinf g x x x= − = −
( ) ( ) ( )sin 3g f x x= −
85.
86.
87. ( ) 220sin(60 ), 0I t t t= π ≥
2 2 1Period: second
60 30Amplitude: 220 220 amperes
T
Aωπ π= = =
π= =
Section 6.4: Graphs of the Sine and Cosine Functions
601
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
88. ( ) 120sin(30 ), 0I t t t= π ≥
2 2 1Period: second
30 15Amplitude: 120 120 amperes
T
Aωπ π= = =
π= =
89. ( ) 220sin(120 )V t t= π
a. Amplitude: 220 220 volts
2 2 1Period: second
120 60
A
Tω
= =π π= = =
π
b, e.
c. V IR=
220sin(120 ) 10
22sin(120 )
( ) 22sin(120 )
t I
t I
I t t
ππ
π
===
d. Amplitude: 22 22 amperesA = =
2 2 1
Period: second120 60
Tωπ π= = =
π
90. ( ) 120sin(120 )V t t= π
a. Amplitude: 120 120 volts
2 2 1Period: second
120 60
A
Tω
= =π π= = =
π
b, e.
c. V IR=
120sin(120 ) 20
6sin(120 )
( ) 6sin(120 )
t I
t I
I t t
π =π =
= π
d. Amplitude: 6 6 amperesA = =
2 2 1
Period: second120 60
Tωπ π= = =
π
91. a. [ ]
( )
( )
( )
2
2
0
2 20
220
( )( )
sin 2
sin 2
sin 2
V tP t
R
V ft
R
V ft
R
Vft
R
=
π =
π=
= π
b. The graph is the reflected cosine graph translated up a distance equivalent to the
amplitude. The period is 1
2 f, so 4 fω = π .
The amplitude is 2 2
0 01
2 2
V V
R R⋅ = .
The equation is:
( ) ( )
( )
2 20 0
20
cos 42 2
1 cos 42
V VP t ft
R R
Vft
R
= − π +
= − π
c. Comparing the formulas:
( ) ( )( )2 1sin 2 1 cos 4
2ft ftπ = − π
92. a. Since the tunnel is in the shape of one-half a sine cycle, the width of the tunnel at its base is one-half the period. Thus,
22(28) 56T
πω
= = = or 28
πω = .
The tunnel has a maximum height of 15 feet so we have 15A = . Using the form
sin( )y A xω= , the equation for the sine
curve that fits the opening is
15sin28
xy
π =
.
b. Since the shoulders are 7 feet wide and the road is 14 feet wide, the edges of the road correspond to 7x = and 21x = .
Chapter 6: Trigonometric Functions
602
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
7 15 215sin 15sin 10.6
28 4 2
π π = = ≈
21 3 15 215sin 15sin 10.6
28 4 2
π π = = ≈
The tunnel is approximately 10.6 feet high at the edge of the road.
93. a. Physical potential: 2
23ω π= ;
Emotional potential: 2
28 14ω π π= = ;
Intellectual potential: 2
33ω π=
b.
00
110
33#1, #2, #3
( )
( )
( )
2#1: 50sin 50
23
#2 : 50sin 5014
2#3 : 50sin 50
33
P t t
P t t
P t t
π = +
π = +
π = +
c. No.
d.
07305
110
7335
#1
#3
#2
Physical potential peaks at 15 days after the
20th birthday, with minimums at the 3rd and 26th days. Emotional potential is 50% at the 17th day, with a maximum at the 10th day and a minimum at the 24th day. Intellectual potential starts fairly high, drops to a minimum at the 13th day, and rises to a maximum at the 29th day.
94. cos , 2 2y x xπ π= − ≤ ≤
95. sin , 2 2y x xπ π= − ≤ ≤
96. Answers may vary.
7 1 1 5 1 13 1
, , , , , , ,6 2 6 2 6 2 6 2
π π π π −
97. Answers may vary.
5 1 1 1 5 1
, , , , , , ,3 2 3 2 3 2 3 2
π π π π − −
98. 2sin 2
sin 1
x
x
= −= −
Answers may vary.
3 7 11
, 2 , , 2 , , 2 , , 22 2 2 2
π π π π − − − − −
99. Answers may vary.
3 5 9
,1 , ,1 , ,1 , ,14 4 4 4
π π π π −
100 – 104. Answers will vary.
105 – 108. Interactive Exercises.
Section 6.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
603
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Section 6.5
1. 4x =
2. True
3. origin; x = odd multiples of 2
π
4. y-axis; x = odd multiples of 2
π
5. cosy x=
6. True
7. The y-intercept of tany x= is 0.
8. coty x= has no y-intercept.
9. The y-intercept of secy x= is 1.
10. cscy x= has no y-intercept.
11. sec 1 when 2 , 0, 2 ;
sec 1 when ,
x x
x x
= = − π π= − = −π π
12. 3
csc 1 when , ;2 2
3csc 1 when ,
2 2
x x
x x
π π= = −
π π= − = −
13. secy x= has vertical asymptotes when
3 3, , ,
2 2 2 2x
π π π π= − − .
14. cscy x= has vertical asymptotes when
2 , , 0, , 2x = − π − π π π .
15. tany x= has vertical asymptotes when
3 3, , ,
2 2 2 2x
π π π π= − − .
16. coty x= has vertical asymptotes when
2 , , 0, , 2x = − π − π π π .
17. 3 tany x= ; The graph of tany x= is stretched
vertically by a factor of 3.
The domain is , is an odd integer2
kx x k
π ≠
.
The range is the set of all real number or ( , )−∞ ∞ .
18. 2 tany x= − ; The graph of tany x= is stretched
vertically by a factor of 2 and reflected about the x-axis.
The domain is , is an odd integer2
kx x k
π ≠
.
The range is the set of all real number or ( , )−∞ ∞ .
19. 4coty x= ; The graph of coty x= is stretched
vertically by a factor of 4.
The domain is { }, is an integerx x k kπ≠ . The
range is the set of all real number or ( , )−∞ ∞ .
20. 3coty x= − ; The graph of coty x= is stretched
vertically by a factor of 3 and reflected about the
Chapter 6: Trigonometric Functions
604
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
x-axis.
The domain is { }, is an integerx x k kπ≠ . The
range is the set of all real number or ( , )−∞ ∞ .
21. tan2
y xπ =
; The graph of tany x= is
horizontally compressed by a factor of 2
π.
The domain is { }does not equal an odd integerx x .
The range is the set of all real number or ( , )−∞ ∞ .
22. 1
tan2
y x =
; The graph of tany x= is
horizontally stretched by a factor of 2.
The domain is { }, is an integerx x k kπ≠ . The
range is the set of all real number or ( , )−∞ ∞ .
23. 1
cot4
y x =
; The graph of coty x= is
horizontally stretched by a factor of 4.
The domain is { }4 , is an integerx x k kπ≠ . The
range is the set of all real number or ( , )−∞ ∞ .
24. cot4
y xπ =
; The graph of coty x= is
horizontally stretched by a factor of 4
π.
The domain is { }4 , is an integerx x k k≠ . The
range is the set of all real number or ( , )−∞ ∞ .
25. 2secy x= ; The graph of secy x= is stretched
vertically by a factor of 2.
The domain is , is an odd integer2
kx x k
π ≠
.
The range is { }2 or 2y y y≤ − ≥ .
Section 6.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
605
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
26. 1
csc2
y x= ; The graph of cscy x= is vertically
compressed by a factor of 1
2.
The domain is { }, is an integerx x k kπ≠ . The
range is 1 1
or 2 2
y y y ≤ − ≥
.
27. 3cscy x= − ; The graph of cscy x= is
vertically stretched by a factor of 3 and reflected about the x-axis.
The domain is { }, is an integerx x k kπ≠ . The
range is { }3 or 3y y y≤ − ≥ .
28. 4secy x= − ; The graph of secy x= is
vertically stretched by a factor of 4 and reflected about the x-axis.
The domain is , is an odd integer2
kx x k
π ≠
.
The range is { }4 or 4y y y≤ − ≥ .
29. 1
4sec2
y x =
; The graph of secy x= is
horizontally stretched by a factor of 2 and vertically stretched by a factor of 4.
The domain is { }, is an odd integerx x k kπ≠ .
The range is { }4 or 4y y y≤ − ≥ .
30. ( )1csc 2
2y x= ; The graph of cscy x= is
horizontally compressed by a factor of 1
2 and
vertically compressed by a factor of 1
2.
The domain is , is an integer2
kx x k
π ≠
. The
range is 1 1
or 2 2
y y y ≤ − ≥
.
Chapter 6: Trigonometric Functions
606
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31. ( )2cscy xπ= − ; The graph of cscy x= is
horizontally compressed by a factor of 1
π,
vertically stretched by a factor of 2, and reflected about the x-axis.
The domain is { }does not equal an integerx x .
The range is { }2 or 2y y y≤ − ≥ .
32. 3sec2
y xπ = −
; The graph of secy x= is
horizontally compressed by a factor of 2
π,
vertically stretched by a factor of 3, and reflected about the x-axis.
The domain is { } does not equal an odd integerx x .
The range is { }3 or 3y y y≤ − ≥ .
33. 1
tan 14
y x = +
; The graph of tany x= is
horizontally stretched by a factor of 4 and shifted
up 1 unit.
The domain is { }2 , is an odd integerx x k kπ≠ .
The range is the set of all real number or ( , )−∞ ∞ .
34. 2cot 1y x= − ; The graph of coty x= is
vertically stretched by a factor of 2 and shifted down 1 unit.
The domain is { }, is an integerx x k kπ≠ . The
range is the set of all real number or ( , )−∞ ∞ .
35. 2
sec 23
y xπ = +
; The graph of secy x= is
horizontally compressed by a factor of 3
2π and
Section 6.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
607
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
shifted up 2 units.
The domain is 3
, is an odd integer4
x x k k ≠
.
The range is { }1 or 3y y y≤ ≥ .
36. 3
csc2
y xπ =
; The graph of cscy x= is
horizontally compressed by a factor of 2
3π.
The domain is 2
, is an integer3
x x k k ≠
. The
range is { }1 or 1y y y≤ − ≥ .
37. 1 1
tan 22 4
y x = −
; The graph of tany x= is
horizontally stretched by a factor of 4, vertically
compressed by a factor of 1
2, and shifted down 2
units.
The domain is { }2 , is an odd integerx x k kπ≠ .
The range is the set of all real number or ( , )−∞ ∞ .
38. 1
3cot 22
y x = −
; The graph of coty x= is
horizontally stretched by a factor of 2, vertically stretched by a factor of 3, and shifted down 2 units.
The domain is { }2 , is an integerx x k kπ≠ . The
range is the set of all real number or ( , )−∞ ∞ .
39. 1
2csc 13
y x = −
; The graph of cscy x= is
horizontally stretched by a factor of 3, vertically stretched by a factor of 2, and shifted down 1 unit.
Chapter 6: Trigonometric Functions
608
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
The domain is { }3 , is an integerx x k kπ≠ .
The range is { }3 or 1y y y≤ − ≥ .
40. 1
3sec 14
y x = +
; The graph of secy x= is
horizontally stretched by a factor of 4, vertically stretched by a factor of 3, and shifted up 1 unit.
The domain is { }2 , is an odd integerx x k kπ≠ .
The range is { }2 or 4y y y≤ − ≥ .
41. ( ) ( ) ( )
30 0tan / 6 tan 06 3/ 6 / 60
6
3 6 2 3
3
f fπ
ππ π π
π π
− − − = =−
= ⋅ =
The average rate of change is 2 3
π.
42. ( ) ( ) ( )
( )
2 30 1sec / 6 sec 06 3/ 6 / 60
6
2 3 2 32 3 3 6
3
f fπ
ππ π π
π π
− − − = =−
−−= ⋅ =
The average rate of change is( )2 3 2 3
π
−.
43. ( ) ( ) ( )0
tan 2 / 6 tan 2 06/ 60
6
3 0 6 3
/ 6
f fπ
ππ π
π π
− ⋅ − ⋅ =−
−= =
The average rate of change is6 3
π.
44. ( ) ( ) ( )0
sec 2 / 6 sec 2 06/ 60
62 1 6
/ 6
f fπ
ππ π
π π
− ⋅ − ⋅ =−
−= =
The average rate of change is6
π.
45. ( )( ) ( )tan 4f g x x=
Section 6.5: Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions
609
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
( ) ( ) ( )4 tan 4 tang f x x x= =
46. ( )( ) 12sec
2f g x x
=
( )( ) ( )12sec sec
2g f x x x= =
47. ( )( ) ( )2 cot 2cotf g x x x= − = −
( )( ) ( )cot 2g f x x= −
48. ( )( ) ( )12csc csc
2f g x x x= =
( )( ) 12csc
2g f x x
=
49.
Chapter 6: Trigonometric Functions
610
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
50.
51. a. Consider the length of the line segment in two sections, x, the portion across the hall that is 3 feet wide and y, the portion across that hall that is 4 feet wide. Then,
3cos
3
cos
x
x
θ
θ
=
=
and 4
sin
4
sin
y
y
θ
θ
=
=
Thus, 3 4
3sec 4csccos sin
L x y θ θθ θ
= + = + = + .
b. Let 1
3 4
cos sinY
x x= + .
00
25
2π_
c. Use MINIMUM to find the least value:
00
25
2π_
L is least when 0.83θ ≈ .
d. ( ) ( )3 4
9.86 feetcos 0.83 sin 0.83
L ≈ + ≈ .
Note that rounding up will result in a ladder
that won’t fit around the corner. Answers will vary.
52. a. ( ) 10 tan( )d t tπ=
b. ( ) 10 tan( )d t tπ= is undefined at 1
2t = and
3
2t = , or in general at
is an odd integer2
kt k =
. At these
instances, the length of the beam of light approaches infinity. It is at these instances in the rotation of the beacon when the beam of light being cast on the wall changes from one side of the beacon to the other.
c. ( ) 10 tan( )
0 0
0.1 3.2492
0.2 7.2654
0.3 13.764
0.4 30.777
t d t tπ=
d. (0.1) (0) 3.2492 0
32.4920.1 0 0.1 0
d d− −= ≈− −
(0.2) (0.1) 7.2654 3.249240.162
0.2 0.1 0.2 0.1
d d− −= ≈− −
(0.3) (0.2) 13.764 7.265464.986
0.3 0.2 0.3 0.2
d d− −= ≈− −
(0.4) (0.3) 30.777 13.764170.13
0.4 0.3 0.4 0.3
d d− −= ≈− −
e. The first differences represent the average rate of change of the beam of light against the wall, measured in feet per second. For example, between 0t = seconds and 0.1t = seconds, the average rate of change of the
Section 6.6: Phase Shift: Sinusoidal Curve Fitting
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
beam of light against the wall is 32.492 feet per second.
53.
Yes, the two functions are equivalent.
Section 6.6
1. phase shift
2. False
3. 4sin(2 )y x= − π
Amplitude: 4 4
2 2Period:
2
Phase Shift:2
A
Tω
φω
= =π π= = = π
π=
Interval defining one cycle:
3, ,
2 2T
φ φ π πω ω + =
Subinterval width:
4 4
T π=
Key points:
,02
π
, 3
, 44
π
, ( ),0π , 5
, 44
π −
, 3
,02
π
4. 3sin(3 )y x= − π
Amplitude: 3 3
2 2Period:
3
Phase Shift:3
A
Tω
φω
= =π π= =
π=
Interval defining one cycle:
, ,3
Tφ φ π πω ω + =
Subinterval width: 2 / 3
4 4 6
T π π= =
Key points:
,03
π
, ,32
π
, 2
,03
π
, 5
, 36
π −
, ( ),0π
Chapter 6: Trigonometric Functions
612
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5. 2cos 32
y xπ = +
Amplitude: 2 2
2 2Period:
3
2Phase Shift:
3 6
A
Tω
φω
= =π π= =
π − π = = −
Interval defining one cycle:
, ,6 2
Tφ φ π πω ω + = −
Subinterval width: 2 / 3
4 4 6
T π π= =
Key points:
, 26
π −
, ( )0,0 , , 26
π −
, ,03
π
, , 22
π
6. ( )3cos 2y x= + π
Amplitude: 3 3
2 2Period:
2
Phase Shift:2 2
A
Tω
φω
= =π π= = = π
−π π= = −
Interval defining one cycle:
, ,2 2
Tφ φ π πω ω + = −
Subinterval width:
4 4
T π=
Key points:
,32
π −
, ,04
π −
, ( )0, 3− , ,04
π
, ,32
π
7. 3sin 22
y xπ = − +
Amplitude: 3 3
2 2Period:
2
2Phase Shift:2 4
A
Tω
φω
= − =π π= = = π
π− π= = −
Interval defining one cycle:
3, ,
4 4T
φ φ π πω ω + = −
Subinterval width:
4 4
T π=
Key points:
,04
π −
, ( )0, 3− , ,04
π
, ,32
π
, 3
,04
π
8. 2cos 22
y xπ = − −
Amplitude: 2 2
2 2Period:
2
2Phase Shift:2 4
A
Tω
φω
= − =π π= = = π
ππ= =
Interval defining one cycle:
Section 6.6: Phase Shift: Sinusoidal Curve Fitting
613
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
5, ,
4 4T
φ φ π πω ω + =
Subinterval width:
4 4
T π=
Key points:
, 24
π −
, ,02
π
, 3
, 24
π
, ( ),0π , 5
, 24
π −
9. 4sin( 2) 5y x= π + −
Amplitude: 4 4
2 2Period: 2
2 2Phase Shift:
A
Tω
φω
= =π π= = =
π−= = −π π
Interval defining one cycle:
2 2, , 2T
φ φω ω π π + = − −
Subinterval width: 2 1
4 4 2
T = =
Key points:
2, 5
π − −
, 1 2
, 12 π
− −
, 2
1 , 5π
− −
,
3 2, 9
2 π − −
, 2
2 , 5π
− −
10. 2cos(2 4) 4y x= π + +
Amplitude: 2 2
2 2Period: 1
24 2
Phase Shift:2
A
Tω
φω
= =π π= = =
π−= = −
π π
Interval defining one cycle:
2 2, ,1T
φ φω ω π π + = − −
Subinterval width: 1
4 4
T =
Key points:
2,6
π −
, 1 2
, 44 π
−
, 1 2
, 22 π
−
, 3 2
, 44 π
−
,
21 ,6
π −
11. 3cos( 2) 5y x= π − +
Amplitude: 3 3
2 2Period: 2
2Phase Shift:
A
Tω
φω
= =π π= = =
π
=π
Interval defining one cycle:
2 2, , 2T
φ φω ω π π + = +
Subinterval width: 2 1
4 4 2
T = =
Key points:
Chapter 6: Trigonometric Functions
614
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
2,8
π
, 1 2
,52 π
+
, 2
1 ,2π
+
, 3 2
,52 π
+
,
22 ,8
π +
12. 2cos(2 4) 1y x= π − −
Amplitude: 2 2
2 2Period: 1
24 2
Phase Shift:2
A
Tω
φω
= =π π= = =
π
= =π π
Interval defining one cycle:
2 2, ,1T
φ φω ω π π + = +
Subinterval width: 1
4 4
T =
Key points:
2,1
π
, 1 2
, 14 π
+ −
, 1 2
, 32 π
+ −
,
3 2, 1
4 π + −
, 2
1 ,1π
+
13. 3sin 2 3sin 22 2
3sin 22
y x x
x
π π = − − + = − − −
π = −
Amplitude: 3 3
2 2Period:
2
2Phase Shift:2 4
A
Tω
φω
= =π π= = = π
ππ= =
Interval defining one cycle:
5, ,
4 4T
φ φ π πω ω + =
Subinterval width:
4 4
T π=
Key points:
,04
π
, ,32
π
, 3
,04
π
, ( ), 3π − , 5
,04
π
14. 3cos 2 3cos 22 2
3cos 22
y x x
x
π π = − − + = − − −
π = − −
Amplitude: 3 3
2 2Period:
2
2Phase Shift:2 4
A
Tω
φω
= − =π π= = = π
ππ= =
Interval defining one cycle:
5, ,
4 4T
φ φ π πω ω + =
Subinterval width:
Section 6.6: Phase Shift: Sinusoidal Curve Fitting
615
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
4 4
T π=
Key points:
, 34
π −
, ,02
π
, 3
,34
π
, ( ),0π , 5
, 34
π −
15. 1
2; ;2
A Tφω
= = π =
2 22
Tω π π= = =
π
1
2 2 1
φ φω
φ
= =
=
Assuming A is positive, we have that sin( ) 2sin(2 1)
1 2sin 2
2
y A x x
x
ω φ= − = −
= −
16. 3; ; 22
A Tφω
π= = =
2 24
2T
ω π π= = =π
24
8
φ φω
φ
= =
=
Assuming A is positive, we have that
( )sin( ) 3sin(4 8)
3sin 4 2
y A x x
x
ω φ= − = −
= −
17. 1
3; 3 ;3
A Tφω
= = π = −
2 2 2
3 3Tω π π= = =
π
12 33
1 2 2
3 3 9
φ φω
φ
= = −
= − ⋅ = −
Assuming A is positive, we have that 2 2
sin( ) 3sin3 9
2 1 3sin
3 3
y A x x
x
ω φ = − = + = +
18. 2; ; 2A Tφω
= = π = −
2 22
Tω π π= = =
π 2
2 4
φ φω
φ
= = −
= −
Assuming A is positive, we have that
( )sin( ) 2sin(2 4)
2sin 2 2
y A x x
x
ω φ= − = +
= +
19. ( )2 tan 4y x π= −
Begin with the graph of tany x= and apply the
following transformations:
1) Shift right π units ( )tany x π = −
2) Horizontally compress by a factor of 1
4
( )tan 4y x π = −
3) Vertically stretch by a factor of 2
( )2 tan 4y x π = −
20. ( )1cot 2
2y x π= −
Begin with the graph of coty x= and apply the
following transformations:
1) Shift right π units ( )coty x π = −
2) Horizontally compress by a factor of 1
2
( )cot 2y x π = −
Chapter 6: Trigonometric Functions
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3) Vertically compress by a factor of 1
2
( )1cot 2
2y x π = −
21. 3csc 24
y xπ = −
Begin with the graph of cscy x= and apply the
following transformations:
1) Shift right 4
π units csc
4y x
π = −
2) Horizontally compress by a factor of 1
2
csc 24
y xπ = −
3) Vertically stretch by a factor of 3
3csc 24
y xπ = −
22. ( )1sec 3
2y x π= −
Begin with the graph of secy x= and apply the
following transformations:
1) Shift right π units ( )secy x π = −
2) Horizontally compress by a factor of 1
3
( )sec 3y x π = −
3) Vertically compress by a factor of 1
2
( )1sec 3
2y x π = −
23. cot 22
y xπ = − +
Begin with the graph of coty x= and apply the
following transformations:
1) Shift left 2
π units cot
2y x
π = +
2) Horizontally compress by a factor of 1
2
cot 22
y xπ = +
3) Reflect about the x-axis
cot 22
y xπ = − +
Section 6.6: Phase Shift: Sinusoidal Curve Fitting
617
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24. tan 32
y xπ = − +
Begin with the graph of tany x= and apply the
following transformations:
1) Shift left 2
π units tan
2y x
π = +
2) Horizontally compress by a factor of 1
3
tan 32
y xπ = +
3) Reflect about the x-axis
tan2
y xπ = − +
,112
π
, 14
π −
25. ( )sec 2y xπ π= − +
Begin with the graph of secy x= and apply the
following transformations:
1) Shift left π units ( )secy x π = +
2) Horizontally compress by a factor of 1
2π
( )sec 2y xπ π = +
3) Reflect about the x-axis
( )sec 2y xπ π = − +
26. 1
csc2 4
y xππ = − − +
Begin with the graph of cscy x= and apply the
following transformations:
1) Shift left 4
π units csc
4y x
π = +
2) Reflect about the y-axis csc4
y xπ = − +
3) Horizontally compress by a factor of 2
π
1csc
2 4y x
ππ = − +
3) Reflect about the x-axis
1csc
2 4y x
ππ = − − +
Chapter 6: Trigonometric Functions
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27. ( ) 120sin 30 , 03
I t t tπ = π − ≥
2 2 1Period: second
30 15Amplitude: 120 120 amperes
13Phase Shift: second30 90
T
Aω
φω
π π= = =π
= =π
= =π
28. ( ) 220sin 60 , 06
I t t tπ = π − ≥
2 2 1Period: second
60 30Amplitude: 220 220 amperes
16Phase Shift: second60 360
T
Aω
φω
π π= = =π
= =π
= =π
29. a.
200
60
13
b. 56.0 24.2 31.8
Amplitude: 15.92 2
A−= = =
56.0+24.2 80.2Vertical Shift: 40.1
2 22
12 6Phase shift (use 24.2, 1):y x
ω
= =
π π= =
= =
24.2 15.9sin 1 40.16
15.9 15.9sin6
1 sin6
2 62
3
φ
φ
φ
φ
φ
π = ⋅ − +
π − = −
π − = −
π π− = −
π=
Thus, 2
15.9sin 40.16 3
y xπ π = − +
or
( )15.9sin 4 40.16
y xπ = − +
.
c.
200
60
13
d. ( )15.62sin 0.517 2.096 40.377y x= − +
e.
200
60
13
30. a.
200
90
13
b. 80.0 34.6 45.4
Amplitude: 22.72 2
A−= = =
80.0+34.6 114.6Vertical Shift: 57.3
2 22
12 6ω
= =
π π= =
Phase shift (use 34.6, 1):y x= =
Section 6.6: Phase Shift: Sinusoidal Curve Fitting
619
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34.6 22.7sin 1 57.36
22.7 22.7sin6
1 sin6
2 62
3
φ
φ
φ
φ
φ
π = ⋅ − +
π − = −
π − = −
π π− = −
π=
Thus, 2
22.7sin 57.36 3
y xπ π = − +
or
( )22.7sin 4 57.36
y xπ = − +
.
c.
200
90
13
d. ( )22.61sin 0.503 2.038 57.17y x= − +
e.
200
90
13
31. a.
200
80
13
b. 75.4 25.5 49.9
Amplitude: 24.952 2
A−= = =
75.4+25.5 100.9Vertical Shift: 50.45
2 22
12 6ω
= =
π π= =
Phase shift (use 25.5, 1):y x= =
25.5 24.95sin 1 50.456
24.95 24.95sin6
1 sin6
2 62
3
φ
φ
φ
φ
φ
π = ⋅ − +
π − = −
π − = −
π π− = −
π=
Thus, 2
24.95sin 50.456 3
y xπ π = − +
or
( )24.95sin 4 50.456
y xπ = − +
.
c.
200
80
13
d. 25.693sin(0.476 1.814) 49.854y x= − +
e.
200
80
13
32. a.
200
80
13
b. 77.0 31.8 45.2
Amplitude: 22.62 2
A−= = =
77.0+31.8 108.8Vertical Shift: 54.4
2 22
12 6ω
= =
π π= =
Phase shift (use 31.8, 1):y x= =
Chapter 6: Trigonometric Functions
620
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31.8 22.6sin 1 54.46
22.6 22.6sin6
1 sin6
2 62
3
φ
φ
φ
φ
φ
π = ⋅ − +
π − = −
π − = −
π π− = −
π=
Thus, 2
22.6sin 54.46 3
y xπ π = − +
or
( )22.6sin 4 54.46
y xπ = − +
.
c.
200
80
13
d. 22.46sin(0.506 2.060) 54.35y x= − +
e.
200
80
13
33. a. 11.5 + 12.4167 = 23.9167 hours which is at 11:55 PM.
b. ( )5.84 0.37 6.21
Ampl: 3.1052 2
A− −
= = =
5.84 ( 0.37) 5.47Vertical Shift: 2.735
2 22 24
12.4167 6.20835 149Phase shift (use 5.84, 11.5):y x
π π πω
+ − = =
= = =
= =
245.84 3.105sin 11.5 2.735
149
243.105 3.105sin 11.5
149
2761 sin
149
276
2 1494.2485
π φ
π φ
π φ
π π φ
φ
= ⋅ − + = ⋅ −
= −
= −
≈
Thus, 24
3.105sin 4.2485 2.735149
y xπ = − +
or
( )243.105sin 8.3958 2.735
149y x
π = − + .
c. ( )243.105sin 15 4.2485 2.735
149
2.12 feet
yπ = − +
≈
34. a. 2.6167 + 12.4167 = 15.0334 hours which is at 3:02 PM.
b. 11.09 ( 2.49) 13.58
Ampl: 6.792 2
A− −= = =
11.09 ( 2.49) 8.6Vertical Shift: 4.3
2 22 24
12.4167 6.20835 149Phase shift (use 11.09, 2.6167):y x
π π πω
+ − = =
= = =
= =
2411.09 6.79sin 2.6167 4.3
149
246.79 6.79sin 2.6167
149
62.80081 sin
149
62.8008
2 1490.2467
π φ
π φ
π φ
π φ
φ
= ⋅ − + = ⋅ −
= −
π = −
≈ −
Thus, 24
6.79sin 0.2467 4.3149
y xπ = + +
or ( )246.79sin 0.4875 4.3
149y x
π = + + .
c. ( )246.79sin 18 0.2467 4.3
149
4.77 feet
yπ = + +
≈
Section 6.6: Phase Shift: Sinusoidal Curve Fitting
621
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35. a. 13.75 10.55
Amplitude: 1.62
A−= =
13.75 10.55Vertical Shift: 12.15
22
365Phase shift (use 13.75, 172):y x
ω
+ =
π=
= =
213.75 1.6sin 172 12.15
365
21.6 1.6sin 172
365
3441 sin
365
344
2 3651.3900
φ
φ
φ
φ
φ
π = ⋅ − +
π = ⋅ −
π = −
π π= −
≈
Thus, 2
1.6sin 1.39 12.15365
y xπ = − +
or
( )21.6sin 80.75 12.15
365y x
π = − + .
b. ( )21.6sin 91 80.75 12.15
365
12.43 hours
yπ = − +
≈
c.
d. The actual hours of sunlight on April 1, 2010 were 12.43 hours. This is the same as the predicted amount.
36. a. 15.30 9.10
Amplitude: 3.12
A−= =
15.30 9.10Vertical Shift: 12.2
22
365Phase shift (use 15.30, 172):y x
ω
+ =
π=
= =
215.30 3.1sin 172 12.2
365
23.1 3.1sin 172
365
3441 sin
365
344
2 3651.39
φ
φ
φ
φ
φ
π = ⋅ − +
π = ⋅ −
π = −
π π= −
≈
Thus, 2
3.1sin 1.39 12.2365
y xπ = − +
or
( )23.1sin 80.75 12.2
365y x
π = − + .
b. ( )23.1sin 91 1.39 12.2
365
12.74 hours
yπ = − +
≈
c.
d. The actual hours of sunlight on April 1, 2010 were 12.72 hours. This is very close to the predicted amount of 12.74 hours.
37. a. 19.42 5.48
Amplitude: 6.972
A−= =
19.42 5.48Vertical Shift: 12.45
22
365Phase shift (use 19.42, 172):y x
ω
+ =
π=
= =
Chapter 6: Trigonometric Functions
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219.42 6.97sin 172 12.45
365
26.975 6.975sin 172
365
3441 sin
365
344
2 3651.39
φ
φ
φ
φ
φ
π = ⋅ − +
π = ⋅ −
π = −
π π= −
≈
Thus, 2
6.97sin 1.39 12.45365
y xπ = − +
or
( )26.97sin 80.75 12.45
365y x
π = − + .
b. ( )26.97sin 91 1.39 12.45
365
13.67 hours
yπ = − +
≈
c.
d. The actual hours of sunlight on April 1, 2010 was 13.42 hours. This is close to the predicted amount of 13.67 hours.
38. a. 13.43 10.85
Amplitude: 1.292
A−= =
13.43 10.85Vertical Shift: 12.14
22
365Phase shift (use 13.43, 172):y x
ω
+ =
π=
= =
213.43 1.29sin 172 12.14
365
21.29 1.29sin 172
365
3441 sin
365
344
2 3651.39
φ
φ
φ
φ
φ
π = ⋅ − +
π = ⋅ − π = −
π π= −
≈
Thus, 2
1.29sin 1.39 12.14365
y xπ = − +
.
b. ( )21.29sin 91 1.39 12.14
365
12.37 hours
yπ = − +
≈
c.
d. The actual hours of sunlight on April 1, 2010 were 12.38 hours. This is very close to the predicted amount of 12.37 hours.
39 – 40. Answers will vary.
Chapter 6 Review Exercises
1. 3
135 135 radian radians180 4
π π° = ⋅ =
2. 7
210 210 radian radians180 6
π π° = ⋅ =
3. 18 18 radian radian180 10
π π° = ⋅ =
4. 15 15 radian radian180 12
π π° = ⋅ =
Chapter 6 Review Exercises
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5. 3 3 180
degrees 1354 4
π π= ⋅ = °π
6. 2 2 180
degrees 1203 3
π π= ⋅ = °π
7. 5 5 180
degrees 4502 2
π π− = − ⋅ = − °π
8. 3 3 180
degrees 2702 2
π π− = − ⋅ = − °π
9. 1 1
tan sin 14 6 2 2
π π− = − =
10. 1 3
cos sin 13 2 2 2
π π+ = + =
11. 2 3 3 2 4 3
3sin 45º 4 tan 3 46 2 3 2 3
π− = ⋅ − ⋅ = −
12. 1
4cos 60º 3tan 4 3 3 2 3 33 2
π+ = ⋅ + ⋅ = +
13. ( )3 26cos 2 tan 6 2 3
4 3 2
3 2 2 3
π π + − = − + −
= − −
14. ( )2 5 3 3 33sin 4cos 3 4 0
3 2 2 2
π π− = − =
15. 5 5
sec cot sec cot 2 1 33 4 3 4
π π π π − − − = + = + =
16. 3 3
4csc cot 4csc cot 4 2 14 4 4 4
π π π π − − = + = +
17. tan sin 0 0 0π + π = + =
18. cos csc cos csc 0 1 12 2 2 2
π π π π − − = + = + =
19. cos540º tan( 405º ) 1 ( 1) 1 1 0− − = − − − = − + =
20. sin 270º cos( 180º ) 1 ( 1) 2+ − = − + − = −
21. 2 2 22
1sin 20º sin 20º cos 20º 1
sec 20º+ = + =
22.
( )
2 22 2
2 2
1 1sec 40º tan 40º
cos 40º cot 40º
1 tan 40º tan 40º
1
− = −
= + −
=
23. 1
sec50º cos50º cos50º 1cos50º
⋅ = ⋅ =
24. 1
tan10º cot10º tan10º 1tan10º
⋅ = ⋅ =
25. sin 50º sin 50º
cos 40º sin(90º 40º )
sin 50º1
sin 50º
=−
= =
26. tan 20º tan 20º
cot 70 tan(90 70 )
tan 20º1
tan 20
=° ° − °
= =°
27. sin( 40º ) sin 40º
cos50º sin(90º 50º )
sin 40º1
sin 40º
− −=−
−= = −
28. ( ) 1tan 20º cot 20º tan 20º 1
tan 20º− = − ⋅ = −
29. ( )
( )
sin 400º sec 50º sin 400º sec50º
1sin 40º 360º
cos50ºsin 40º sin 40º
cos50º sin(90º 50º )
sin 40º1
sin 40º
⋅ − = ⋅
= + ⋅
= =−
= =
30. ( )cot 200º cot 70º cot 200º ( cot 70º )
cot 200º cot 70º cot(20º 180º ) cot 70º
sin 20º sin 70º
cos 20º cos 70ºsin 20º sin 70º
sin(90º 20º ) sin(90º 70º )
sin 20º sin 70º1
sin 70º sin 20º
⋅ − = ⋅ −= − ⋅ = − + ⋅
= − ⋅
= − ⋅− −
= − ⋅ = −
Chapter 6: Trigonometric Functions
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31. 4
sin5
θ = and 02
πθ< < , so θ lies in quadrant I.
Using the Pythagorean Identities: 2 2
22
cos 1 sin
4 16 9cos 1 1
5 25 25
9 3cos
25 5
θ θ
θ
θ
= −
= − = − =
= ± = ±
Note that cosθ must be positive since θ lies in
quadrant I. Thus, 3
cos5
θ = .
45
35
sin 4 5 4tan
cos 5 3 3
θθθ
== = ⋅ =
45
1 1 5 5csc 1
sin 4 4θ
θ== = ⋅ =
35
1 1 5 5sec 1
cos 3 3θ
θ== = ⋅ =
43
11 1 3 3
cottan 4 4
θθ
== = ⋅ =
32. 1
tan4
θ = and 02
πθ< < , so θ lies in quadrant I.
Using the Pythagorean Identities: 2 2
22
sec tan 1
1 1 17sec 1 1
4 16 16
17 17sec
16 4
θ θ
θ
θ
= +
= + = + =
= ± = ±
Note that secθ must be positive since θ lies in
quadrant I. Thus, 17
sec4
θ = .
174
1 1cos
sec
17 4 4 17 4 171
4 1717 17 17
θθ
= =
= ⋅ = = ⋅ =
sintan
cos
θθθ
= , so
( )( ) 1 4 17 17sin tan cos
4 17 17θ θ θ
= = =
1717
1 1 17csc 1
sin 17
17 17 17 17 1717
1717 17 17
θθ
= = = ⋅
= = ⋅ = =
14
1 1cot 4
tanθ
θ= = =
33. 12
tan5
θ = and sin 0θ < , so θ lies in quadrant III.
Using the Pythagorean Identities: 2 2
22
sec tan 1
12 144 169sec 1 1
5 25 25
169 13sec
25 5
θ θ
θ
θ
= +
= + = + =
= ± = ±
Note that secθ must be negative since θ lies in
quadrant III. Thus, 13
sec5
θ = − .
135
1 1 5cos
sec 13θ
θ= = = −
−
sintan
cos
θθθ
= , so
( )( ) 12 5 12sin tan cos
5 13 13θ θ θ = = − = −
1213
1 1 13csc
sin 12θ
θ= = = −
−
125
1 1 5cot
tan 12θ
θ= = =
34. 12
cot5
θ = and cos 0θ < , so θ lies in quadrant III.
Using the Pythagorean Identities: 2 2
22
csc 1 cot
12 144 169csc 1 1
5 25 25
169 13csc
25 5
θ θ
θ
θ
= +
= + = + =
= ± = ±
Note that cscθ must be negative because θ lies in
quadrant III. Thus, 13
csc5
θ = − .
135
1 1 5sin
csc 13θ
θ= = = −
−
coscot
sin
θθθ
= , so
Chapter 6 Review Exercises
625
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( ) ( ) 12 5 12cos cot sin
5 13 13θ θ θ = = − = −
125
1 1 5tan
cot 12θ
θ= = =
1213
1 1 13sec
cos 12θ
θ= = = −
−
35. 5
sec4
θ = − and tan 0θ < , so θ lies in quadrant II.
Using the Pythagorean Identities: 2 2
22
tan sec 1
5 25 9tan 1 1
4 16 16
9 3tan
16 4
θ θ
θ
θ
= −
= − − = − =
= ± = ±
Note that tan 0θ < , so 3
tan4
θ = − .
54
1 1 4cos
sec 5θ
θ= = = −
−
sintan
cos
θθθ
= , so
( )( ) 3 4 3sin tan cos
4 5 5θ θ θ = = − − =
.
35
1 1 5csc
sin 3θ
θ= = =
34
1 1 4cot
tan 3θ
θ= = = −
−
36. 5
csc3
θ = − and cot 0θ < , so θ lies in quadrant
IV. Using the Pythagorean Identities:
2 2
22
cot csc 1
5 25 16cot 1 1
3 9 9
16 4cot
9 3
θ θ
θ
θ
= −
= − − = − =
= ± = ±
Note that cot 0θ < , so 4
cot3
θ = − .
53
1 1 3sin
csc 5θ
θ= = = −
−
coscot
sin
θθθ
= , so
( ) ( ) 4 3 4cos cot sin
3 5 5θ θ θ = = − − =
43
1 1 3tan
cot 4θ
θ= = = −
−
45
1 1 5sec
cos 4θ
θ= = =
37. 12
sin13
θ = and θ lies in quadrant II.
Using the Pythagorean Identities: 2 2
22
cos 1 sin
12 144 25cos 1 1
13 169 169
25 5cos
169 13
θ θ
θ
θ
= −
= − = − =
= ± = ±
Note that cosθ must be negative because θ lies
in quadrant II. Thus, 5
cos13
θ = − .
1213
513
sin 12 13 12tan
cos 13 5 5
θθθ
= = = − = − −
1213
1 1 13csc
sin 12θ
θ= = =
513
1 1 13sec
cos 5θ
θ= = = −
−
125
1 1 5cot
tan 12θ
θ= = = −
−
38. 3
cos5
θ = − and θ lies in quadrant III.
Using the Pythagorean Identities: 2 2
22
sin 1 cos
3 9 16sin 1 1
5 25 25
16 4sin
25 5
θ θ
θ
θ
= −
= − − = − =
= ± = ±
Note that sinθ must be negative because θ lies
in quadrant III. Thus, 4
sin5
θ = − .
45
35
sin 4 5 4tan
cos 5 3 3
θθθ
− = = = − − = −
45
1 1 5csc
sin 4θ
θ= = = −
−
Chapter 6: Trigonometric Functions
626
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
35
1 1 5sec
cos 3θ
θ= = = −
−
43
1 1 3cot
tan 4θ
θ= = =
39. 5
sin13
θ = − and 3
22
π θ< < π (quadrant IV)
Using the Pythagorean Identities: 2 2
22
cos 1 sin
5 25 144cos 1 1
13 169 169
144 12cos
169 13
θ θ
θ
θ
= −
= − − = − =
= ± = ±
Note that cosθ must be positive because θ lies
in quadrant IV. Thus, 12
cos13
θ = .
513
1213
sin 5 13 5tan
cos 13 12 12
θθθ
− = = = − = −
513
1 1 13csc
sin 5θ
θ= = = −
−
1213
1 1 13sec
cos 12θ
θ= = =
512
1 1 12cot
tan 5θ
θ= = = −
−
40. 12
cos13
θ = and 3
22
π θ< < π (quadrant IV)
Using the Pythagorean Identities: 2 2
22
sin 1 cos
12 144 25sin 1 1
13 169 169
25 5sin
169 13
θ θ
θ
θ
= −
= − = − =
= ± = ±
Note that sinθ must be negative because θ lies
in quadrant IV. Thus, 5
sin13
θ = − .
513
1213
sin 5 13 5tan
cos 13 12 12
θθθ
− = = = − = −
513
1 1 13csc
sin 5θ
θ= = = −
−
1213
1 1 13sec
cos 12θ
θ= = =
512
1 1 12cot
tan 5θ
θ= = = −
−
41. 1
tan3
θ = and 180º 270ºθ< < (quadrant III)
Using the Pythagorean Identities: 2 2
22
sec tan 1
1 1 10sec 1 1
3 9 9
10 10sec
9 3
θ θ
θ
θ
= +
= + = + =
= ± = ±
Note that secθ must be negative since θ lies in
quadrant III. Thus, 10
sec3
θ = − .
1 1 3 10 3 10cos
sec 1010 10 103
θθ
= = = − ⋅ = −−
sintan
cos
θθθ
= , so
( )( ) 1 3 10 10sin tan cos
3 10 10θ θ θ
= = − = −
1 1 10csc 10
sin 10 1010
θθ
= = = − = −−
13
1 1cot 3
tanθ
θ= = =
42. 2
tan3
θ = − and 90º 180ºθ< < (quadrant II)
Using the Pythagorean Identities: 2 2
22
sec tan 1
2 4 13sec 1 1
3 9 9
13 13sec
9 3
θ θ
θ
θ
= +
= − + = + =
= ± = ±
Note that secθ must be negative since θ lies in
quadrant II. Thus, 13
sec3
θ = − .
1 1 3 13 3 13cos
sec 1313 13 133
θθ
= = = − ⋅ = −−
Chapter 6 Review Exercises
627
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
sintan
cos
θθθ
= , so
( )( ) 2 3 13 2 13sin tan cos
3 13 13θ θ θ
= = − − =
1 1 13 13csc
sin 22 13 2 1313
θθ
= = = =
23
1 1 3cot
tan 2θ
θ= = = −
−
43. sec 3θ = and 3
22
π θ< < π (quadrant IV)
Using the Pythagorean Identities: 2 2
2 2
tan sec 1
tan 3 1 9 1 8
tan 8 2 2
θ θθθ
= −
= − = − =
= ± = ±
Note that tanθ must be negative since θ lies in
quadrant IV. Thus,. tan 2 2θ = − . 1 1
cossec 3
θθ
= =
sintan
cos
θθθ
= , so
( )( ) 1 2 2sin tan cos 2 2
3 3θ θ θ = = − = −
.
2 23
1 1 3 2 3 2csc
sin 2 42 2θ
θ= = = − ⋅ = −
−
1 1 2 2cot
tan 42 2 2θ
θ= = ⋅ = −
−
44. csc 4θ = − and 3
2
πθπ < < (quadrant III)
Using the Pythagorean Identities:
( )
2 2
22
cot csc 1
cot 4 1 16 1 15
cot 15
θ θ
θ
θ
= −
= − − = − =
= ±
Note that cotθ must be positive since θ lies in
quadrant III. Thus, cot 15θ = . 1 1 1
sincsc 4 4
θθ
= = = −−
coscot
sin
θθθ
= , so
( ) ( ) 1 15cos cot sin 15
4 4θ θ θ = = − = −
1 1 15 15tan
cot 1515 15θ
θ= = ⋅ =
154
1 1 4 15 4 15sec
cos 1515 15θ
θ= = = − ⋅ = −
−
45. cot 2θ = − and 2
π θ< < π (quadrant II)
Using the Pythagorean Identities:
( )
2 2
22
csc 1 cot
csc 1 2 1 4 5
csc 5
θ θ
θ
θ
= +
= + − = + =
= ±
Note that cscθ must be positive because θ lies
in quadrant II. Thus, csc 5θ = .
1 1 5 5sin
csc 55 5θ
θ= = ⋅ =
coscot
sin
θθθ
= , so
( ) ( ) 5 2 5cos cot sin 2
5 5θ θ θ
= = − = −
.
1 1 1tan
cot 2 2θ
θ= = = −
−
2 55
1 1 5 5sec
cos 22 5θ
θ= = = − = −
−
46. tan 2θ = − and 3
22
π θ< < π (quadrant IV)
Using the Pythagorean Identities:
( )
2 2
22
sec tan 1
sec 2 1 4 1 5
sec 5
θ θ
θ
θ
= +
= − + = + =
= ±
Note that secθ must be positive since θ lies in
quadrant IV. Thus, sec 5θ = .
1 1 5 5cos
sec 55 5θ
θ= = ⋅ =
sintan
cos
θθθ
= , so
( ) ( ) 5 2 5sin tan cos 2
5 5θ θ θ
= = − = −
.
Chapter 6: Trigonometric Functions
628
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1 1 5 5csc
sin 22 5 2 55
θθ
= = = − = −−
1 1 1cot
tan 2 2θ
θ= = = −
−
47. 2sin(4 )y x=
The graph of siny x= is stretched vertically by
a factor of 2 and compressed horizontally by a
factor of 1
4.
Domain: ( ),−∞ ∞
Range: [ ]2,2−
48. 3cos(2 )y x= −
The graph of cosy x= is stretched vertically by
a factor of 3, reflected across the x-axis, and
compressed horizontally by a factor of 1
2.
Domain: ( ),−∞ ∞
Range: [ ]3,3−
49. 2cos2
y xπ = − +
The graph of cosy x= is shifted 2
π units to the
left, stretched vertically by a factor of 2, and reflected across the x-axis.
Domain: ( ),−∞ ∞
Range: [ ]2,2−
50. 3sin( )y x= − π
The graph of siny x= is shifted π units to the
right, and stretched vertically by a factor of 3.
Domain: ( ),−∞ ∞
Range: [ ]3,3−
Chapter 6 Review Exercises
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51. tan( )y x= + π
The graph of tany x= is shifted π units to the
left.
Domain: | , k is an odd integer2
kx x
π ≠
Range: ( ),−∞ ∞
52. tan2
y xπ = − −
The graph of tany x= is shifted 2
π units to the
right and reflected across the x-axis.
Domain: { }| , k is an integerx x kπ≠
Range: ( ),−∞ ∞
53. 2 tan(3 )y x= −
The graph of tany x= is stretched vertically by
a factor of 2, reflected across the x-axis, and
compressed horizontally by a factor of 1
3.
Domain: | , k is an integer6 3
x x kπ π ≠ + ⋅
Range: ( ),−∞ ∞
54. 4 tan(2 )y x=
The graph of tany x= is stretched vertically by
a factor of 4 and compressed horizontally by a
factor of 1
2.
Domain: | , k is an odd integer4
kx x
π ≠
Range: ( ),−∞ ∞
Chapter 6: Trigonometric Functions
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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
55. cot4
y xπ = +
The graph of coty x= is shifted 4
π units to the
left.
Domain: | , k is an integer4
x x kπ π ≠ − +
Range: ( ),−∞ ∞
56. 4cot(2 )y x= −
The graph of coty x= is stretched vertically by
a factor of 4, reflected across the x-axis and
compressed horizontally by a factor of 1
2.
Domain: | , k is an integer2
kx x
π ≠
Range: ( ),−∞ ∞
57. ( )4sec 2y x=
The graph of secy x= is stretched vertically by
a factor of 4 and compressed horizontally by a
factor of 1
2.
4321
�2�3�4�5
x
y
�( , �4)�––2
( , �4)�––2
(0, 4) (�, 4)(��, 4)
�––4
3�–––4
�––4�
3�–––4�
Domain: | , k is an odd integer4
kx x
π ≠
Range: { }| 4 or 4y y y≤ − ≥
58. csc4
y xπ = +
The graph of cscy x= is shifted 4
π units to the
left.
Domain: | , k is an integer4
x x kπ π ≠ − +
Range: { }| 1 or 1y y y≤ − ≥
59. ( )4sin 2 4 2y x= + −
The graph of siny x= is shifted left 4 units,
compressed horizontally by a factor of 1
2,
stretched vertically by a factor of 4, and shifted
Chapter 6 Review Exercises
631
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
down 2 units.
Domain: ( ),−∞ ∞
Range: [ ]6,2−
60. ( )3cos 4 2 1y x= + +
The graph of cosy x= is shifted left 2 units,
compressed horizontally by a factor of 1
4,
stretched vertically by a factor of 3, and shifted up 1 unit.
Domain: ( ),−∞ ∞
Range: [ ]2,4−
61. 4 tan2 4
xy
π = +
The graph of tany x= is stretched horizontally
by a factor of 2, shifted left 4π units, and
stretched vertically by a factor of 4.
Domain: | 2 , k is an integer2
x x kπ π ≠ + ⋅
Range: ( ),−∞ ∞
62. 5cot3 4
xy
π = −
The graph of coty x= is shifted right 4π units,
stretched horizontally by a factor of 3, and stretched vertically by a factor of 5.
Domain: 3
| 3 ,k is an integer4
x x kπ π ≠ + ⋅
Range: ( ),−∞ ∞
63. 4cosy x=
Amplitude = 4 4= ; Period = 2π
64. sin(2 )y x=
Amplitude = 1 1= ; Period = 2
2
π π=
65. 8sin2
y xπ = −
Amplitude = 8 8− = ; Period = 2
4
2
ππ =
Chapter 6: Trigonometric Functions
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66. 2cos(3 )y x= − π
Amplitude = 2 2− = ; Period = 2 2
3 3
ππ
=
67. 4sin(3 )y x=
Amplitude: 4 4
2 2Period:
30
Phase Shift: 03
A
Tω
φω
= =π π= =
= =
68. 1
2cos3
y x =
Amplitude: 2 2
2 2Period: 6
13
0Phase Shift: 0
13
A
Tω
φω
= =π π= = = π
= =
69. ( )2sin 2y x π= −
Amplitude: 2 2
2 2Period:
2
Phase Shift:2 2
A
T πω
φ π πω
= =π π= = =
= =
70. 1
cos2 2
y xπ = − +
Amplitude: 1 1
2 2Period: 4
12
2Phase Shift:12
A
T πω
πφ πω
= − =π π= = =
−= = −
71. 1 3
sin2 2
y x = − π
1 1Amplitude:
2 2
2 2 4Period:
3 32
2Phase Shift:
3 32
A
Tω
φω
= =
π π π= = =
π π= =
Chapter 6 Review Exercises
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72. ( )3cos 6 3
2y x= + π
3 3Amplitude:
2 2
2 2Period:
6 33
Phase Shift:6 2
A
Tω
φω
= =
π π π= = =
− π π= = −
73. ( )2cos 6
3y x= − π −
2 2Amplitude:
3 3
2 2Period: 2
6Phase Shift:
A
Tω
φω
= − =
π π= = =π
=π
74. 4
7sin3 3
y xπ = − −
Amplitude: 7 7
2 2Period: 6
34
43Phase Shift:
3
A
Tω
φω
= − =π π= = =
π
= =π π
75. The graph is a cosine graph with amplitude 5 and
period 8π.
Find ω : 2
8
8 2
2 1
8 4
ωω
ω
ππ =
π = ππ= =π
The equation is: 1
5cos4
y x =
.
76. The graph is a sine graph with amplitude 4 and period 8π.
Find ω : 2
8
8 2
2 1
8 4
ωω
ω
ππ =
π = ππ= =π
The equation is: 1
4sin4
y x =
.
77. The graph is a reflected cosine graph with amplitude 6 and period 8.
Chapter 6: Trigonometric Functions
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Find ω : 2
8
8 2
2
8 4
ωω
ω
π=
= ππ π= =
The equation is: 6cos4
y xπ = −
.
78. The graph is a reflected sine graph with amplitude 7 and period 8.
Find ω : 2
8
8 2
2
8 4
ωω
ω
π=
= ππ π= =
The equation is: 7sin4
y xπ = −
.
79. Set the calculator to radian mode: sin 0.388
π ≈ .
80. Set the calculator to degree mode:
oo
1sec10 1.02
cos10= ≈ .
81. Terminal side of θ in Quadrant III implies sin 0 csc 0
cos 0 sec 0
tan 0 cot 0
θ θθ θθ θ
< << <> >
82. cos 0, tan 0θ θ> < ; θ lies in quadrant IV.
83. 1 2 2
,3 3
P
= −
2 2sin
3t = ;
1 3 2 3 2csc
42 2 22 23
t = = ⋅ =
1cos
3t = − ;
1sec 3
13
t = = − −
2 22 2 33tan 2 2
3 113
t
= = ⋅ − = − −
;
1 2 2cot
42 2 2t = ⋅ = −
−
84. The point ( 2, 5)P = − is on a circle of radius
2 2( 2) 5 4 25 29r = − + = + = with the center
at the origin. So, we have 2x = − , 5y = , and
29r = . Thus, 5 5 29
sin2929
yt
r= = = ;
2 2 29cos
2929
xt
r
−= = = − ; 5
tan2
yt
x= = − .
85. The domain of secy x= is
odd multiple of 2
x xπ ≠
.
The range of secy x= is { }1 or 1 .y y y≤ − ≥
The period is 2π .
86. a. o o20 3532 20 '35" 32 32.34
60 3600= + + ≈
b. o63.18 o0.18 (0.18)(60 ') 10.8'
0.8' (0.8)(60") 48"
= == =
Thus, o o63.18 63 10 '48"=
87. 2 feet, 30º or 6
r θ θ π= = =
( )22
2 1.047 feet6 3
1 12 1.047 square feet
2 2 6 3
s r
A r
θ
θ
π π= = ⋅ = ≈
π π= ⋅ = ⋅ ⋅ = ≈
88. In 30 minutes: 8 inches, 180º or r θ θ= = = π
8 8 25.13 inchess rθ= = ⋅ π = π ≈
In 20 minutes: 2
8 inches, 120º or 3
r θ θ π= = =
2 168 16.76 inches
3 3s rθ π π= = ⋅ = ≈
Chapter 6 Review Exercises
635
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
89. 180 mi/hrv = ; 1
mile21
0.25 mile4
d
r
=
= =
180 mi/hr
0.25 mi720 rad/hr
720 rad 1 rev
hr 2 rad360 rev
hr114.6 rev/hr
v
rω = =
=
= ⋅π
=π
≈
90. 25 feetr = ; 1 rev 1 rev 2 radians
rad/sec30 sec 30 sec 1 rev 15
ω π π= = ⋅ =
525 ft/sec 5.24 ft/sec.
15 3v rω π π= = ⋅ = ≈
The linear speed is approximately 5.24 feet per
second; the angular speed is 1
30 revolutions per
second, or 15
π radians per second.
91. Since there are two lights on opposite sides and the light is seen every 5 seconds, the beacon makes 1 revolution every 10 seconds:
1 rev 2 radians radians/second
10 sec 1 rev 5ω π π= ⋅ =
92. 16 inches; 90 mi/hrr v= =
90 mi/hr 12 in 5280 ft 1 hr 1 rev
16 in 1 ft 1 mi 60 min 2 rad945.38 rev/min
v
rω =
= ⋅ ⋅ ⋅ ⋅π
≈
Yes, the setting will be different for a wheel of radius 14 inches:
14 inches; 90 mi/hrr v= =
90 mi/hr 12 in 5280 ft 1 hr 1 rev
14 in 1 ft 1 mi 60 min 2 rad1080.43 rev/min
v
rω =
= ⋅ ⋅ ⋅ ⋅π
≈
93. ( )( ) 120sin 120 , 0E t t t= π ≥
a. The maximum value of E is the amplitude, which is 120.
b. 2 1
Period120 60
π= =π
sec.
c.
94. ( ) 220sin 30 , 06
I t t tπ = π + ≥
a. 2 1
Period30 15
π= =π
b. The amplitude is 220.
c. The phase shift is:
1 1630 6 30 180
φω
π− π= = − ⋅ = −π π
d.
95. a. 100
50
90
80
70
60
12961 3 52 4 7 8 10 110
y
x
b. 90 51 39
Amplitude: 19.52 2
A−= = =
90+51 141Vertical Shift: 70.5
2 22
12 6ω
= =
π π= =
Phase shift (use 51, 1):y x= =
Chapter 6: Trigonometric Functions
636
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51 19.5sin 1 70.56
19.5 19.5sin6
1 sin6
2 62
3
φ
φ
φ
φ
φ
π = ⋅ − +
π − = −
π − = −
π π− = −
π=
Thus, 2
19.5sin 70.56 3
y xπ π = − +
, or
( )19.5sin 4 70.56
y xπ = − +
.
c. 100
50
90
80
70
60
12961 3 52 4 7 8 10 110
y
x
d. ( )19.52sin 0.54 2.28 71.01y x= − +
e.
450
95
13
96. a. 14.63 9.72
Amplitude: 2.4552
A−= =
14.63 9.72Vertical Shift: 12.175
22
365ω
+ =
π=
Phase shift (use 14.63, 172):y x= =
214.63 2.455sin 172 12.175
365
22.455 2.455sin 172
365
φ
φ
π = ⋅ − +
π = ⋅ −
3441 sin
365344
2 3651.39
φ
φ
φ
π = −
π π= −
≈
Thus, 2
2.455sin 1.39 12.175365
y xπ = − +
,
or 2
2.455sin ( 80.75) 12.175365
y xπ = − +
.
b. 2
2.455sin (91) 1.39 12.175365
12.61 hours
yπ = − +
≈
c.
20
y
x
10
0 140 280 420
d. The actual hours of sunlight on April 1, 2005 was 12.62 hours. This is close to the predicted amount of 12.61 hours.
Chapter 6 Test
637
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
97.
Chapter 6 Test
1. 260 260 1 degree
260 radian180
260 13 radian radian
180 9
π
π π
° = ⋅
= ⋅
= =
2. 400 400 1 degree
400 radian180
400 20 radian radian
180 9
π
π π
− ° = − ⋅
= − ⋅
= − = −
3. 13
13 13 1 degree 13 radian radian180 180π π° = ⋅ = ⋅ =
4. radian 1 radian8 8
180degrees 22.5
8
π π
ππ
− = − ⋅
= − ⋅ = − °
5. 9 9
radian 1 radian2 2
9 180degrees 810
2
π π
ππ
= ⋅
= ⋅ = °
6. 3 3
radian 1 radian4 4
3 180 degrees 135
4
π π
ππ
= ⋅
= ⋅ = °
7. 1
sin6 2π =
8. 5 3 5 3os cos os cos
4 4 4 4
3 3cos cos
4 4
c c 2
0
π π π π
π π
π− − = − −
− =
+
=
9. ( ) ( ) 1cos 120 cos 120
2− ° = ° = −
10. ( ) ( ) 3tan 330 tan 150 180 tan 150
3° = ° + ° = ° = −
11.
( )
19 3sin tan sin tan 4
2 4 2 4
3sin tan 1 1 2
2 4
π π π π π
π π
− = − + = − = − − =
12.
( )
22 3 2
2sin 60 3cos 45 2 32 2
3 1 23 3 2 3 3 22
4 2 2 2 2
° − ° = −
− = − = − =
13. Set the calculator to degree mode: sin17 0.292° ≈
14. Set the calculator to radian mode: 2
cos 0.3095π ≈
15. Set the calculator to degree mode: 1
sec 229 1.524cos 229
° = ≈ −°
Chapter 6: Trigonometric Functions
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16. Set the calculator to radian mode: 28 1
cot 2.747289 tan
9
ππ= ≈
17. To remember the sign of each trig function, we primarily need to remember that sinθ is positive in quadrants I and II, while cosθ is positive in quadrants I and IV. The sign of the other four trig functions can be determined directly from sine and
cosine by knowing sin
tancos
θθ θ= , 1
seccos
θ θ= ,
1csc
sinθ θ= , and
coscot
sinθθ θ= .
sin cos tan sec csc cot
in QI
in QII +
in QIII +
in QIV +
θ θ θ θ θ θθθθθ
+ + + + + +− − − + −
− − − − +− − + − −
18. Because ( ) sinf x x= is an odd function and
since 3
( ) sin5
f a a= = , then
3( ) sin( ) sin
5f a a a− = − = − = − .
19. 5
sin7
θ = and θ in quadrant II.
Using the Pythagorean Identities: 2
2 2 5 25 24cos 1 sin 1 1
7 49 49
24 2 6cos
49 7
θ θ
θ
= − = − = − =
= ± = ±
Note that cosθ must be negative because θ lies
in quadrant II. Thus, 2 6
cos7
θ = − .
57
2 67
sin 5 7 6 5 6tan
cos 7 122 6 6
θθθ
= = = − ⋅ = − −
57
1 1 7csc
sin 5θ
θ= = =
2 67
1 1 7 6 7 6sec
cos 122 6 6θ
θ= = = − ⋅ = −
−
5 612
1 1 12 6 2 6cot
tan 55 6 6θ
θ= = = − ⋅ = −
−
20. 2
cos3
θ = and 3
22π θ π< < (in quadrant IV).
Using the Pythagorean Identities: 2
2 2 2 4 5sin 1 cos 1 1
3 9 9
5 5sin
9 3
θ θ
θ
= − = − = − =
= ± = ±
Note that sinθ must be negative because θ lies
in quadrant IV. Thus, 5
sin3
θ = − .
53
23
sin 5 3 5tan
cos 3 2 2
θθθ
−= = = − ⋅ = −
53
1 1 3 5 3 5csc
sin 55 5θ
θ= = = − ⋅ = −
−
23
1 1 3sec
cos 2θ
θ= = =
52
1 1 2 5 2 5cot
tan 55 5θ
θ= = = − ⋅ = −
−
21. 12
tan5
θ = − and 2π θ π< < (in quadrant II)
Using the Pythagorean Identities: 2
2 2 12 144 169sec tan 1 1 1
5 25 25
169 13sec
25 5
θ θ
θ
= + = − + = + =
= ± = ±
Note that secθ must be negative since θ lies in
quadrant II. Thus, 13
sec5
θ = − .
135
1 1 5cos
sec 13θ
θ= = = −
−
sintan
cos
θθθ
= , so
( )( ) 12 5 12sin tan cos
5 13 13θ θ θ = = − − =
Chapter 6 Test
639
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1213
1 1 13csc
sin 12θ
θ= = =
125
1 1 5cot
tan 12θ
θ= = = −
−
22. The point ( )2,7 lies in quadrant I with 2x =
and 7y = . Since 2 2 2x y r+ = , we have
2 22 7 53r = + = . So,
7 7 53 7 53sin
5353 53 53
yr
θ = = = ⋅ = .
23. The point ( )5,11− lies in quadrant II with
5x = − and 11y = . Since 2 2 2x y r+ = , we
have ( )2 25 11 146r = − + = . So,
5 5 146 5 146cos
146146 146 146
xr
θ − −= = = ⋅ = − .
24. The point ( )6, 3− lies in quadrant IV with 6x =
and 3y = − . Since 2 2 2x y r+ = , we have
( )226 3 45 3 5r = + − = = . So,
3 1tan
6 2yx
θ −= = = −
25. Comparing 2sin3 6
xy
π = −
to
( )siny A xω φ= − , we see that
2A = , 1
3ω = , and
6
πφ = . The graph is a sine
curve with amplitude 2A = , period
2 26
1/ 3T
π π πω
= = = , and phase shift
6
1/ 3 2
πφ πω
= = = . The graph of 2sin3 6
xy
π = −
will lie between 2− and 2 on the y-axis. One
period will begin at 2
xφ πω
= = and end at
2 136
2 2x
π φ π ππω ω
= + = + = . We divide the
interval 13
,2 2
π π
into four subintervals, each of
length 6 3
4 2
π π= .
7 7 13,2 , 2 , , ,5 , 5 ,
2 2 2 2
π π π ππ π π π
The five key points on the graph are
( ) ( )7 13,0 , 2 ,2 , ,0 , 5 , 2 , ,0
2 2 2
π π ππ π −
We plot these five points and fill in the graph of the sine function. The graph can then be extended in both directions.
321 x
y
(2�, 2)(�4�, 2)
(��, �2)
(5�, �2)0�5�–––2 (( ,0�
11�––––2 (( ,
07�–––2 (( ,
�––2 (( , 0
26. tan 24
y xπ = − + +
Begin with the graph of tany x= , and shift it
4
π units to the left to obtain the graph of
tan4
y xπ = +
. Next, reflect this graph about
the y-axis to obtain the graph of tan4
y xπ = − +
.
Finally, shift the graph up 2 units to obtain the
graph of tan 24
y xπ = − + +
.
y
x−π
2
π− π2
tan 24
y xπ = − + +
4
Chapter 6: Trigonometric Functions
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27. For a sinusoidal graph of the form
( )siny A xω φ= − , the amplitude is given by
A , the period is given by 2πω , and the phase
shift is given by φω . Therefore, we have 3A = − ,
3ω = , and 3
34 4π πφ = − = −
. The equation
for the graph is 3
3sin 34
y xπ = − +
.
28. The area of the walk is the difference between the area of the larger sector and the area of the smaller shaded sector.
Wa
lk
50°
3 ft The area of the walk is given by
( )2 2
2 2
1 12 2
2
A R r
R r
θ θθ
= −
= −
,
where R is the radius of the larger sector and r is the radius of the smaller sector. The larger radius is 3 feet longer than the smaller radius because the walk is to be 3 feet wide. Therefore,
3R r= + , and
( )( )( )( )
2 2
2 2
32
6 92
6 92
A r r
r r r
r
θ
θ
θ
= + −
= + + −
= +
The shaded sector has an arc length of 25 feet
and a central angle of 5
50 radians18π° = . The
radius of this sector is 518
25 90 feet
sr πθ π
= = = .
Thus, the area of the walk is given by 518
2 2
90 5 5406 9 9
2 365
75 ft 78.93 ft4
Aπ π
π ππ
= + = +
= + ≈
29. To throw the hammer 83.19 meters, we need 2
0
20
2
2 2 20
0
83.19 m9.8 m/s815.262 m /28.553 m/s
vs
g
v
v sv
=
=
==
Linear speed and angular speed are related according to the formula v r ω= ⋅ . The radius is
190 cm 1.9 mr = = . Thus, we have
( )28.55328.553 1.9
15.028 radians per secondradians 60 sec 1 revolution
15.028sec 1 min 2 radians
143.5 revolutions per minute (rpm)
r ωω
ω
ω π
= ⋅==
= ⋅ ⋅
≈
To throw the hammer 83.19 meters, Adrian must have been swinging it at a rate of 143.5 rpm upon release.
Chapter 6 Cumulative Review
1.
( )( )22 1 0
2 1 1 0
x x
x x
+ − =− + =
1 or 1
2x x= = −
The solution set is 1
1,2
−
.
2. Slope 3= − , containing (–2,5)
1 1Using ( )y y m x x− = −
( )5 3 ( 2)
5 3( 2)5 3 6
3 1
y x
y xy x
y x
− = − − −− = − +− = − −
= − −
3. radius = 4, center (0,–2)
( ) ( )2 2 2Using x h y k r− + − =
( ) ( )( )( )
22 2
22
0 2 4
2 16
x y
x y
− + − − =
+ + =
4. 2 3 12x y− =
This equation yields a line.
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2 3 12
3 2 12
24
3
x y
y x
y x
− =− = − +
= −
The slope is 2
3m = and the y-intercept is 4− .
Let 0y = : 2 3(0) 12
2 12
6
x
x
x
− ===
The x-intercept is 6.
5. 2 2 2 4 4 0x y x y+ − + − =
( ) ( )( ) ( )
2 2
2 2
2 2 2
2 1 4 4 4 1 4
1 2 9
1 2 3
x x y y
x y
x y
− + + + + = + +
− + + =
− + + =
This equation yields a circle with radius 3 and center (1,–2).
6. 2( 3) 2y x= − +
Using the graph of 2y x= , horizontally shift to
the right 3 units, and vertically shift up 2 units.
7. a. 2y x=
b. 3y x=
c. xy e=
Chapter 6: Trigonometric Functions
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d. lny x=
e. siny x=
f. tany x=
8. ( ) 3 2f x x= −
( )1
3 2
3 2 Inverse
2 3
2
32 1
( ) 23 3
y x
x y
x y
xy
xf x x−
= −= −
+ =+ =
+= = +
9. ( ) ( )2 2o osin14 cos14 3 1 3 2+ − = − = −
10. 3sin(2 )y x=
Amplitude: 3 3
2Period:
20
Phase Shift: 02
A
T
φω
= =π= = π
= =
11. 3
tan 3cos csc 1 3 24 6 6 2
3 33
2
6 3 3
2
π π π − + = − +
= −
−=
12. We need a function of the form xy A b= ⋅ , with
0, 1b b> ≠ . The graph contains the points
( ) ( )0,2 and 1,6 . Therefore, 02
2 1
2
A b
A
A
= ⋅= ⋅=
.
And 1
2
6 2
3
xy b
b
b
=
==
So we have the function 2 3xy = ⋅ .
13. The graph is a cosine graph with amplitude 3 and period 12.
Find ω : 2
12
12 2
2
12 6
ωω
πω
π=
= ππ= =
The equation is: 3cos6
y xπ =
.
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14. a. Given points ( 2, 3)− and (1, 6)− , we
compute the slope as follows:
( )2 1
2 1
6 3 9slope 3
1 2 3
y y
x x
− − − −= = = = −− − −
1 1Using ( )y y m x x− = − :
( )( )( )
3 3 2
3 3 2
3 6 33 3
y x
y x
y xy x
− = − − −− = − +
= − − += − −
The linear function is ( ) 3 3f x x= − − .
Slope: 3m = − ;
y-intercept: ( ) ( )0 3 0 3 3f = − − = −
x-intercept: 0 3 33 3
1
xxx
= − −= −= −
Intercepts: ( )1,0− , ( )0, 3−
b. Given that the graph of ( ) 2f x ax bx c= + +
has vertex (1, –6) and passes through the
point (–2,3), we can conclude 12
b
a− = ,
( )2 3f − = , and ( )1 6f = − .
Notice that 12
2
b
ab a
− =
= −
Also note that
( )( ) ( )2
2 3
2 2 3
4 2 3
f
a b c
a b c
− =
− + − + =− + =
( )( ) ( )2
1 6
1 1 6
6
f
a b c
a b c
= −
+ + = −+ + = −
Replacing b with 2a− in these equations
yields: ( )4 2 2 33 8
a a c
c a
− − + == −
and 2 66
a a cc a
− + = −= − +
So 3 8 69 9
1
a aa
a
− = − +==
Thus,
( )22 1 2
b a
b
= −= − = −
and
( )3 83 8 1 5
c a
c
= −= − = −
Therefore, we have the function
( ) ( )22 2 5 1 6f x x x x= − − = − − .
y-intercept: ( ) ( )20 0 2 0 5 5f = − − = −
x-intercepts: 20 2 5x x= − −
( ) ( ) ( )( )( )
22 2 4 1 5
2 1
2 4 20 2 24 2 2 6
2 2 21 6 1.45 or 3.45
x− − ± − − −
=
± + ± ±= = =
= ± ≈ −
Intercepts: ( )0, 5− , ( )1 6,0− , ( )1 6,0+
c. If ( ) xf x ae= contains the points (–2,3) and
(1,–6), we would have the equations
( ) 22 3f ae−− = = and ( ) 11 6f ae= = − .
Note that 2
22
3
33
ae
a ee
−
−
=
= =
But 1 6
6
ae
ae
= −
= −
Since 2 63e
e≠ − , there is no exponential
function of the form ( ) xf x ae= that
contains the points (–2,3) and (1,–6).
Chapter 6: Trigonometric Functions
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15. a. A polynomial function of degree 3 whose x-intercepts are –2, 3, and 5 will have the form
( ) ( 2)( 3)( 5)f x a x x x= + − − , since the x-
intercepts correspond to the zeros of the function. Given a y-intercept of 5, we have
(0) 5(0 2)(0 3)(0 5) 5
30 55 1
30 6
fa
a
a
=+ − − =
=
= =
Therefore, we have the function 1
( ) ( 2)( 3)( 5)6
f x x x x= + − − .
b. A rational function whose x-intercepts are 2− , 3, and 5 and that has the line 2x = as a
vertical asymptote will have the form ( 2)( 3)( 5)
( )2
a x x xf x
x
+ − −=−
, since the x-
intercepts correspond to the zeros of the numerator, and the vertical asymptote corresponds to the zero of the denominator. Given a y-intercept of 5, we have
(0) 5(0 2)(0 3)(0 5)
50 2
30 101
3
fa
a
a
=+ − − =
−= −
= −
Therefore, we have the function
2
3 2
3 2 3 2
1( 2)( 3)( 5)
3( )2
( 2)( 8 15)
3( 2)
( 6 30)
3( 2)
6 30 6 30
3 6 6 3
x x xf x
xx x x
x
x x x
x
x x x x x x
x x
− + − −=
−+ − +=
− −− − +=− −
− − + − − += =− + −
_________________________________________________________________________________
Chapter 6 Projects
Project I – Internet-Based Project
Project II
1. November 15: High tide: 11:18 am and 11:15 pm November 19: low tide: 7:17 am and 8:38 pm
2. The low tide was below sea level. It is measured against calm water at sea level.
3. Nov Low Tide Low Tide High Tide High Tide
Time Ht (ft) t Time Ht (ft) t Time Ht (ft) t Time Ht (ft) t 14
0-24 6:26a 2.0 6.43 4:38p 1.4 16.63 9:29a 2.2 9.48 11:14p 2.8 23.23
15 24-48
6:22a 1.6 30.37 5:34p 1.8 41.57 11:18a 2.4 35.3 11:15p 2.6 47.25
16 6:28a 1.2 54.47 6:25p 2.0 66.42 12:37p 2.6 60.62 11:16p 2.6 71.27
Chapter 6 Projects
645
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48-72 17
72-96 6:40a 0.8 78.67 7:12p 2.4 91.2 1:38p 2.8 85.63 11:16p 2.6 95.27
18 96-120
6:56a 0.4 102.93 7:57p 2.6 115.95 2:27p 3.0 110.45 11:14p 2.8 119.23
19 120-144
7:17a 0.0 127.28 8:38p 2.6 140.63 3:10p 3.2 135.17 11:05p 2.8 143.08
20 144-168
7:43a -0.2 151.72 3:52p 3.4 159.87
4. The data seems to take on a sinusoidal shape (oscillates). The period is approximately 12 hours. The amplitude varies each day: Nov 14: 0.1, 0.7 Nov 15: 0.4, 0.4 Nov 16: 0.7, 0.3 Nov 17: 1.0, 0.1 Nov 18: 1.3, 0.1 Nov 19: 1.6, 0.1 Nov 20: 1.8
5. Average of the amplitudes: 0.66 Period : 12 Average of vertical shifts: 2.15 (approximately) There is no phase shift. However, keeping in mind the vertical shift, the amplitude
( )siny A Bx D= +
0.66A = 2
12
0.526
B
B
π
π
=
= ≈
2.15D =
Thus, ( )0.66sin 0.52 2.15y x= +
(Answers may vary)
6. ( )0.848sin 0.52 1.25 2.23y x= + +
The two functions are not the same, but they are similar.
7. Find the high and low tides on November 21 which are the min and max that lie between
168t = and 192t = . Looking at the graph of
the equation for part (5) and using MAX/MIN for values between 168t = and 192t = :
Low tides of 1.49 feet when t = 178.2 and
t = 190.3.
High tides of 2.81 feet occur when t = 172.2 and t = 184.3.
Looking at the graph for the equation in part (6) and using MAX/MIN for values between t = 168 and t = 192:
A low tide of 1.38 feet occurs when t = 175.7 and 187.8t = .
A high tide of 3.08 feet occurs when t = 169.8 and 181.9t = .
Chapter 6: Trigonometric Functions
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8. The low and high tides vary because of the moon phase. The moon has a gravitational pull on the water on Earth.
Project III
1. ( )0( ) 1sin 2s t f tπ=
2. 00 0
2 1
2T
f f
ππ
= =
3. 0 0 0 0
1 1 3 10
4 2 4
( ) 0 1 0 1 0
tf f f f
s t −
4. Let 0 1f = =1. Let 0 12x≤ ≤ , with 0.5xΔ = .
Label the graph as 00 12x T≤ ≤ , and each tick
mark is at 0
1
2x
fΔ = .
−2
0
2
12
00
1212T
f= __
5. 0
1
4t
f= ,
0
5
4t
f= ,
0
9
4t
f= , … ,
0
45
4t
f=
6. M = 0 1 0 → P = 0 π 0
7. 0 0( ) 1sin(2 0)S t f tπ= + , 1 0( ) 1sin(2 )S t f tπ π= +
8. [0, 4 0T ] 0S
[4 0T , 8 0T ] 1S
[ 8 0T , 12 0T ] 0S
−2
0
2
12
Project IV
1. Lanai:
3960
mi
θ
s
Oahu
Lanai
3960
mi
θ
s = 65 mi
Oahu
Lanai
3,370 ft
3960 mi
}}
Peak of Lanaihale
2.
650.0164
3960
s r
s
r
θ
θ
=
= = =
3. 3960
cos(0.164)3960
3960 0.9999(3960 )
0.396 miles
0.396 5280 2090 feet
hh
h
=+
= +=
× =
4. Maui:
Oahu
Maui
3960
mi
θ
s=110 mi
Oahu
Maui
10,023 ft
3960 mi
}}
Peak of Haleakala
3960
mi
θ
s
110
0.02783960
s
rθ = = =
3960cos(0.278)
39603960 0.9996(3960 )
1.584 miles
1.584 5280 8364 feet
hh
h
h
=+
= +== × =
Hawaii:
Chapter 6 Projects
647
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall
Oahu
Hawaii
3960
mi
θ
s=190 mi
Oahu
Hawaii
13,796 ft
3960 mi
}}
Peak of Mauna Kea
3960
mi
θ
s
190
0.04803960
s
rθ = = =
3960cos(0.480)
39603960 0.9988(3960 )
4.752 miles
4.752 5280 25,091 feet
hh
h
h
=+
= +== × =
Molokai:
Oahu
Molokai
3960
mi
θs = 40 m
i
Oahu
Molokai
4,961 ft
3960 mi
}}
Peak of Kamakou
3960
mi
θ
s
40
0.01013960
s
rθ = = =
3960cos(0.0101)
39603960 0.9999(3960 )
0.346 miles
0.346 5280 2090 feet
hh
h
h x
=+
= +== =
5. Kamakou, Haleakala, and Lanaihale are all visible from Oahu.
Project V
Answers will vary.